m 


''■r-r: 


'.fe*::^ 


M 


w 


rS^ 


?^*';.'?ii 


my^ 


■•'it  "r^-  ■ 


L/Cl^iyC     C^ .   /(Ji^^.-'^C-v^.a^-#--»-«--<^L_, 


-f-  "/T 


THE 

ELEMENTS  OF  EUCLID, 

VIZ. 

THE  FIRST  SIX  BOOKS, 

TOGETHER   WITH 

THE  ELEVENTH  AND  TWELFTH 

THE  ERRORS, 

»Y  WHICH    THEON,    OR    OTHERS,    HAVE    LONG    AGO    VITIATED    THESE 
BOOKS,    ARE    CORRECTED, 

AND  SOME  OF  EUCLID'S  DEMONSTRATIONS  ARE  RESTORED. 

ALSO,    THE 

BOOK  OF  EUCLID'S  DATA, 

IN  LIKE  MANNER  CORRECTED. 


BY  ROBERT  SIMSON,  M.  D. 

EMERITUS   PROFESSOR    OF    MATHEMATICS    IN    THE    UNIVERSITY 
OF    GLASGOW. 


TO    THIS    EDITION    ARE    ALSO    ANNE.XED, 

ELfiMENTS  OF  PLANE  AND  SPHERICAL  TRIGONOMETRY. 


PHILADELPHIA : 

PUBLISHED    BY    MATHEW   CAREY, 

AND    SOLD  BY   J.    CONRAD    &    CO.    S.   F.    BRADFORD,    BIRCH    &    SMALL, 
AND    SAMUEL    ETHERIDGE. 

yRINTED  BY  T.  ISf  G.  PALMER,  116,  HIGH-STR»«T. 

1806. 


TO  THE  KING, 

THIS  EDITION 

OF 

THE  PRINCIPAL  BOOKS 

or    THE 

ELEMENTS  OF  EUCLID, 

AND    OF    THE 

BOOK  OF  HIS  DATAy 

IS  MOST  JIUMBLY  DEDICATED 
BY 

HIS  MAJESTY'S 

MOST    DUTIFUL, 

AND   MOST   DEVOTED 

SUBJECT   AND    SERVANT 

ROBERT  SIMSON. 


PREFACE. 


The  opinions  of  the  modems  concerning  the  author  of  the 
Elements  of  Geometry,  which  go  under  Euclid's  name,  are 
very  different  and  contrary  to  one  another.  Peter  Ramus 
ascribes  the  propositions,  as  well  as  their  demonstrations,  to 
Theon ;  others  think  the  propositions  to  be  Euclid's,  but  that 
the  demonstrations  are  Theon's  ;  and  others  maintain  that  all 
the  propositions  and  their  demonstrations  are  Euclid's  own. 
John  Buteo  and  Sir  Henry  Savile  are  the  authors  of  greatest 
note  who  assert  this  last,  and  the  greater  part  of  geometers  have 
ever  since  been  of  this  opinion,  as  they  thought  it  the  most 
probable.  Sir  Henry  Savile,  after  the  several  arguments  he 
brings  to  prove  it,  makes  this  conclusion  (page  1 3  Praelect.), 
"  That,  excepting  a  very  few  interpolations,  explications,  and 
"  additions,  Theon  altered  nothing  in  Euclid."  But,  by  often, 
considering  and  comparing  together  the  definitions  and  de- 
monstrations as  they  are  in  the  Greek  editions  we  now  have, 
I  found  that  Theon,  or  whoever  was  the  editor  of  the  present 
Greek  text,  by  adding  some  things,  suppressing  others,  and 
mixing  his  own  with  Euclid's  demonstrations,  had  changed 
more  things  to  the  worse  than  is  commonly  supposed,  and 
those  not  of  small  moment,  especially  in  the  fifth  and  eleventh 
books  of  the  Elements,  which  this  editor  has  greatly  vitiated ; 
for  instance,  by  substituting  a  shorter,  but  insufficient  demon- 
stration of  the  18th  prop,  of  the  5th  book,  in  place  of  the  legiti- 
mate one  which  Euclid  had  given ;  and  by  taking  out  of  this 
book,  besides  other  things,  the  good  definition  which  Eudoxus 
or  Euclid  had  given  of  compound  ratio,  and  given  an  absurd  one 
in  place  of  it  in  the  5th  definition  of  the  6th  book,  which  nei- 
ther Euclid,  Archimedes,  Appolonius,  nor  any  geometer  be- 
fore Theon's  time,  ever  made  use  of,  and  of  which  there  is 
not  to  be  found  the  least  appearance  in  any  of  their  writings ; 


THE 

ELEMENTS  OF  EUCLID. 


BOOK  I. 


/  DEFINITIONS. 

^  '  I.  Book  I. 

A  POINT  is  that  which  hath  no  parts,  or  which  hath  no  mag- 1  ^^ 
nitude.  N?tes. 

11. 
A  line  is  length  without  breadth. 

III. 
The  extremities  of  a  line  are  points. 

IV. 
A  straight  line  is  that  which  lies  evenly  between  its  extreme 
points. 

V. 
A  supei'ficies  is  that  which  hath  only  length  and  breadth. 

VI. 
The  extremities  of  a  superficies  are  lines. 

VII. 
A  plane  superficies  is  that  in  which  any  two  points  being  taken,  See  N. 
the  straight  line  between  them  lies  wholly  in  that  superficies. 
VIII. 
"  A  plane  angle  is  the  inclination  of  two  lines  to  one  another  See  N, 
"  in  a  plane,  which  meet  together,  but  are  not  in  the  same 
"  direction." 

IX. 
A  plane  rectilineal  angle  is  the  inclination  of  two  straight  lines 
to  one  another,  which  meet  together,  but  are  not  in  the  same 
straight  line. 

B 


10 

Book  I. 


THE  ELEMENTS 


B 


N.  B.  '  When  several  angles  are  at  .one  point  B,  any  one  of 

*  them  is  expressed  by  three  letters,  of  which  the  letter  that  is 

*  at  the  vertex  of  the  angle,  that  is,  at  the  point  in  which  the 
'  straight  lines  that  contain  the  angle  meet  one  another,  is  put 

*  between  the  other  two  letters,  and  one  of  these  two  is  some- 
'  where  upon  one  of  those  straight  lines,  and  the  other  upon 
'the  other  line:  thus  the  angle  which  is  contained  by  the 
'  straight  lines  AB,  CB  is  named,  the  angle  ABC,  or  CBA ;  that 
'  vv'hich  is  contained  by  AB,  DB  is  named  the  angle  ADB,  or 
'  DBA ;  and  that  which  is  contained  by  DB,  CB  is  called  the 

*  angle  DBC,  or  CBD  ;  but,  if  there  be  only  one  angle  at  a 
'  point,  it  may  be  expressed  by  a  letter  placed  at  that  point ;  as 
<  the  angle  at  E.' 

X. 

When  a  straight  line  standing  on  ano- 
ther straight  lirhe  makes  the  adjacent 
angles  equal  to  one  another,  each  of 
the  angles  is  called  a  right  angle'; 
and  the  straight  line  which  stands  on 
the  other  is  called  a  perpendicular  to 

it.  

XI. 

^n  obtuse  angle  is  that  v/hich  is  greater  than  a  right  angle. 


XII. 

An  acute  angle  is  that  which  is  less  than  a  right  anp-le. 

XIII. 
"  A  term  or  boundary  is  the  extremity  of  any  thing." 

XIV. 
A  figure  is  th^it  vvhich  is  enclosed  by  onQ  or  more  boundaries. 


OF  EUCLID.  11 

XV.  Book  h 

A  circle  is  a  plane  figure  contained  by  one  line,  which  is  called       »  -^ 
the  circumference,  and  is  such  that  all  straight  lines  drawn 
from  a  certain  point  within  the  figure  to  the  circumferenccj 
are  equal  to  one  another : 


XVI. 

And  this  point  is  called  the  centre  of  the  circle. 

XVII. 

A  diameter  of  a  circle  is  a  straight  line  drawn  through  the  cen- 
tre, and  terminated  both  wdys  by  the  circumference. 
XVIII. 

A  semicircle  is  the  figure  contained  by  a  diameter  and  the  part 
of  the  circuniference  cut  off  by  the  diameter. 

XIX. 

"  A  segment  of  a  circle  is  the  figure  contained  by  a  straight  line 
"  and  the  circumference  it  cuts  off." 

XX. 

Rectilineal  figures  are  those  which  are  contained  by  straight 
lines. 

XXI. 

Trilateral  figures,  or  triangles,  by  three  straight  lines. 

XXII. 
Quadrilateral,  by  four  straight  lines. 

XXIII. 

Multilateral  figures,  or  polygons,  by  more  than  four  straight 
lines. 

XXIV. 

Of  three  sided  figures,  an  equilatei'al  triangle  is  that  which  has 
three  equal  sides. 

XXV. 

An  isosceles  triangle  is  that  which  has  only  two  sides  equal. 


THE  ELEMENTS 


XXVI. 

A  scalene  triangle  is  that  which  has  three  unequal  sides. 

XXVII. 
A  right  angled  triangle  is  that  which  has  a  right  angle. 

XXVIII. 
An  obtuse  angled  triangle  is  that  which  has  an  obtuse  angle. 


XXIX. 

An  acute  angled  triangle  is  that  which  has  three  acute  angles. 

XXX. 
Of  four  sided  figures,  a  square  is  that  which  has  all  its  sides 
tqual,  and  all  its  angles  right  angles. 


XXXI. 

An  oblong  is  that  which  has  all  its  angles  right  angles,  but  has 

not  all  its  sides  equal. 

XXXII. 
A  rhombus  is  that  which  has  all  its  sides  equal,  but  its  angles  are 

not  right  angles. 


XXXIII. 

A  rhomboid  is  that  which  has  its  opposite  sides  equal  to  one 
another,''but  all  its  sides  are  not  equal,  nor  its  angles  right 
angles. 


OF  EUCLID* 

XXXIV. 

All  other  four  sided  figures,  besides  these,  are  called  trape- 
ziums. 

XXXV. 

Parallel  straight  lines  are  such  as  are  in  the  same  plane,  and 
which  being  produced  ever  so  far  both  ways,  do  not  meet. 


POSTULATES. 


I. 

LET  it  be  granted  that  a  straight  line  may  be  drawn  from  any 
one  point  to  any  other  point. 

II. 
That  a  terminated  straight  line  may  be  produced  to  any  length 
in  a  straight  line. 

III. 
And  that  a  circle  may  be  described  from  any  centre,  at  any 
distance  from  that  centre. 


AXIOMS. 

I. 
THINGS  which  are  equal  to  the  same  are  equal  to  one  an- 
other. 

II. 
If  equals  be  added  to  equals,  the  wholes  are  equal. 

III. 
If  equals  be  taken  from  equals,  the  remainders  are  equal. 

IV. 
If  equals  be  added  to  unequals,  the  wholes  are  unequal. 

V. 
If  equals  be  taken  from  unequals,  the  remainders  are  unequal. 

VL 
Things  which  are  double  of  the  same  are  equal  to  one  another. 

VII. 
Things  which  are  halves  of  the  same  are  equal  to  one  another. 

VIII. 
Magnitudes  which  coincide  with  one  another,  that  is,  which 
exactly  fill  the  same  space,  are  equal  to  one  another. 


14  THE  ELEMENTS 

Book  I.  IX. 

^— v*^  The  whole  is  greater  than  its  part. 

X. 

Two  straight  lines  cannot  enclose  a  space. 

XL 

All  right  angles  are  equal  to  one  another. 

XII. 

"  If  a  straight  line  meets  two  straight  lines,  so  as  to  make  the 
"  two  interior  angles  on  the  same  side  of  it  taken  together 
"  less  than  two  right  angles,  these  straight  lines  being  con- 
"  tinually  produced,  shall  at  length  meet  upon  that  side  on 
"  which  are  the  angles  which  are  less  than  two  right  angles. 
"  See  the  notes  on  prop.  29.  of  book  I." 


OF  EUCLID. 

PROPOSITION  I.    PROBLEM. 

TO  describe  an  equilateral  triangle  upon  a  given 
finite  straight  line. 

Let  AB  be  the  given  straight  line ;   it  is  required  to  describe 
an  equilateral  triangle  upon  it. 

C 

From  the  centre  A,  at  the  dis-  

tance    AB,    describe  *    the    circle .      y"^       /^)^\       ^^N.        ^  ^-  ^°^' 
BCD,  and  from  the  centre  B,  at     /  //  \\  \      tulate. 

the  distance  BA,  describe  the  cir- 
cle ACE ;  and  from  the  point  C, 
in  which  the  circles  cut  one  ano- 
ther, draw  the  straight  lines '^  C  A,  \  \  /I  bl.Post. 
CB  to  the  points  A,  B  ;  ABC  shall 
be  an  equilateral  triangle. 

Because  the  point  A  is  the  centre  of  the  circle  BCD,  AC  is 
equal  <=  to  AB ;    and  because  the  point  B  is  the  centre  of  thee  15.  De- 
circle  ACF,  BC  is  equal  to  BA :  but  it  has  been  proved  that  CA  finition. 
is  equal  to  AB ;    therefore  CA,  CB  are  each  of  them  equal  to 
AB ;    but  things  which  are  equal  to  the  same  are  equal  to  one 
another  d  ;  therefore  CA  is  equal  to  CB  ;  wherefore  CA,  AB,  BC  d  1st  Ax- 
are  equal  to  one  another ;    and  the  triangle    ABC  is  therefore  ^°'"* 
equilateral,  and  it  is  described  upon  the  given  straight  line  AB. 
Which  was  required  to  be  done. 

PROP.  II.    PROB. 

FROM  a  given  point  to  draw  a  straight  line  equal 
to  a  given  straight  line. 

Let  A  be  the  given  point,  and  BC  the  given  straight  line ;  it  is 
required  to  draw  from  the  point  A  a  straight  line  equal  to  BC. 


From  the  point  A  to  B  draw* 
the  straight  line  AB ;  and  upon  it 
describe  ^  the  equilateral  triangle 
DAB,  and  produce  «  the  straight 
lines  D  A,  DB  to  E  and  F ;  from' 
the  centre  B,  at  the  distance  BC, 
describe  "^  the  circle  CGH,  and 
from  the  centre  D,  at  the  distance 
DG,  describe  the  circle  GKL.  AL 
chall  be  equal  to  BC. 


a  1.  Post. 

bl.  1.     . 

c  2.  Post, 


d  3.  Post, 


16  THE  ELEMENTS 

Book  I.       Because  the  point  B  is  the  centi-e  of  the  circle  CGH,  BC  is 

*— V— '  equal  e  to  BGj   and  because  D  is  the  centre  of  the  ciixle  GKL, 

elS.Def.  DL  is  equal  to  DG,  and  DA,  DB,  parts  of  them,  are  equal; 

i  3.  Ax.   therefore  the  remainder  AL  is  equal   to  the  remainder*"  BG: 

but  it  has  been  shown,  that  BC  is  equal  to  BG ;    wherefore  AL 

and  BC  are  each  of  them  equal  to  BG ;    and  things  that  are 

equal  to  the  same   are   equal   to  one   another ;     therefore  the 

straight  line  AL  is  equal  to  BC.      Wherefore  from  the  given 

point  A  a  straight  line  AL  has  been  drawn  equal  to  the  given 

straight  line  BC.     Which  was  to  be  done. 

PROP.  in.     PROB. 

FROM  the  greater  of  two  given  straight  lines  to 
cut  off  a  part  equal  to  the  less. 

Let  AB  and  C  be  the  two  given 
straight  lines,  whereof  AB  is  the 
greater.  It  is  required  to  cutoff 
from  AB,  the  greater,  a  partj 
equal  to  C,  the  less. 

a  2. 1.  From  the  point  A  draw  ^  the 

straight  line  AD  equal  to  C  ;  and 
from  the  centre  A,  and  at  the  dis- 

b  3.  Post,  tance  AD,  describe  ^  the   circle 

DEF  ;  and  because  A  is  the  cen-  F 

tre  of  the  circle  DEF,  AE  shall  be  equal  to  AD  ;  but  the  straight 
line  C  is  likewise  equal  to  AD  ;  whence  AE  and  C  are  each  of 
them  equal  to  AD:  wherefore  the  straight  line  AE  is  equal  to*^ 
C,  and  from  AB,  the  greater  of  two  straight  lines,  a  part  AE  has 
been  cut  off  equal  to  C  the  less.     Which  was  to  he  done. 

PROP.  IV.     THEOREM. 

IF  two  triangles  have  two  sides  of  the  one  equal 
to  two  sides  of  the  other,  each  to  each ;  and  have 
likewise  the  angles  contained  by  those  sides  equal 
to  one  another ;  they  shall  likewise  have  their  bases, 
or  third  sides,  equal ;  and  the  two  triangles  shall  be 
equal ;  and  their  other  angles  shall  be  equal,  each  to 
each,  viz.  those  to  which  the  equal  sides  are  opposite. 

Let  ABC,  DEF  be  two  triangles  which  have  the  two  sides 
AB,  AC  equal  to  the  two  sides  DE,  DF,  each  to  each,  viz. 


c  1.  Ax. 


OF  EUCLID.  17 

AB  to  DE,  and  AC  to  DF ;      A  D  Book  I. 

and  the  angle  BAG  equal  to 

the  angle  EDF,  the  base  BC 

shall  be   equal    to   the    base 

EF ;   and  the  triangle   ABC 

to   the    triangle    DEF ;    and 

the  other   angles,  to   which 

the   equal  sides  are  opposite, 

shall  be  equal,  each  to  each, 

viz.  the   angle  ABC  to  the  B 

angle  DEF,  and   the    angle 

ACB  to  DFE. 

For,  if  the  triangle  ABC  be  applied  to  DEF,  so  that  the  point 
A  may  be  on  D,  and  the  straight  line  AB  upon  DE ;  the  point 
B  shall  coincide  with  the  point  E,  because  AB  is  equaj  to  DE ; 
and  AB  coinciding  with  DE,  AC  shall  coincide  with  DF,  be- 
cause the  angle  BAC  is  equal  to  the  angle  EDF;  wherefore 
also  the  point  C  shall  coincide  with  the  point  F,  because  the 
straight  line  AC  is  equal  to  DF :  but  the  point  B  coincides  with 
the  point  E;  wherefore  the  base  BC  shall  coincide  with  the  base 
EF,  because  the  point  B  coinciding  with  E,  and  C  Avith  F,  if 
the  base  BC  does  not  coincide  with  the  base  EF,  two  straight 
lines  would  inclose  a  space,  which  is  impossible^.  Therefore  a  10.  Axi 
the  base  BC  shall  coincide  with  the  base  EF,  and  be  equal  to 
it.  Wherefore  the  whole  triangle  ABC  shall  coincide  with  the 
whole  triangle  DEF,  and  be  equal  to  it ;  and  the  other  angles 
of  the  one  shall  coincide  with  the  remaining  angles  of  the  other, 
and  be  equal  to  them,  viz.  the  angle  ABC  to  the  angle  DEF, 
and  the  angle  ACB  to  DFE.  Therefore,  if  two  triangles  have 
two  sides  of  the  one  equal  to  two  sides  of  the  other,  each  to 
each,  and  have  likewise  the  angles  contained  by  those  sides 
equal  to  one  another,  their  bases  shall  likewise  be  equal,  and  the 
triangles  be  equal,  and  their  other  angles  to  which  the  equal 
sides  are  opposite  shall  be  equal,  each  to  each.  Which  was  to 
be  demonstrated. 


PROP.  V.     THEOR. 

THE  angles  at  the  base  of  an  isosceles  triangle  are 
equal  to  one  another ;  and,  if  the  equal  sides  be  pro- 
duced, the  angles  upon  the  other  side  of  the  base 
shall  be  equal. 

Let  ABC  be  an  ir,osceIes  triangle,  of  which  the  side  AB  ip 

C 


18 


THE  ELEMENTS 


Book  I.  equal  to  AC,  and  let  the  straight  lines  AB,  AC  be  produced  to 
^— y*— ■>.  D  and  E ;  the  angle  ABC  shall  be  equal  to  the  angle  ACB,  and 
the  ongle  CBD  to  the  angle  BCE. 

In  BD  take  any  point  F,  and  ffom  AE  the  greater,  cut  off  AG 
a  3. 1.      equals  to  AF,  the  less,  and  join  FC,  GB. 

Because  AF  is  equal  to  AG,   and  AB   to   AC,  the   two  sides 
FA,  AC  are  equal  to  the  two  GA,   AB,  each  to  each;  and 
they    contain    the    angle    FAG    conj- 
mon     to     the     two     triangles     AFC, 
AGB ;     therefore     the    base     FC     is 
b  4. 1.      equal  ^  to   the    base    GB,   and   the  tri- 
angle   AFC    to    the    triangle    AGB; 
and  the   remaining  angles  of  the  one 
are    equal  •^  to    the    remaining    angles 
of  the   other,  each  to  each,  to  which 
the    equal    sides    are    opposite ;     viz. 
the   angle   ACF   to   the   angle  ABG, 
and    the    angle    AFC    to    the    angle 
AGB :    and   because    the    whole    AF 
is  equal  to   the   whole  AG,  of  which 
the   parts    AB,    AC    are    equal ;     the 
c  3.  Ax.  remainder  BF  shall  be  equal  =  to  the  remainder  CG ;   and  FC 
Avas  proved  to  be  equal  to  GB  ;   therefore  the  two  sides  BF,  FC 
are   equal  to  the  two  CG,   GB,   each   to   each  ;   and   the  angle 
BFC  is  equal  to  the  ?ngle  CGB,  and  the  base  BC  is  common  to 
the  two  triangles  BFC,  CGB  ;  wherefore  the  triangles  are  equal  W 
and  their  remaining  angles,  each  to  each,  to  which  the   equal 
sides  are  opposite  ;  therefore  tlie  angle  FBC  is  equal  to  the  angle 
GCB,  and  the  angle  BCF  to  the  angle  CBG  :  and,  since  it  has 
been  demonstrated  that  the  whole  angle  ABG  is  equal  to  the 
whole  ACF,  the  parts  of  which,  the  angles  CBG,  BCF,  are  also 
equal ;  the  remaining  angle  ABC  is  therefore  equal  to  the  re- 
nnaining  angle  ACB,  v/hicii   are   the  angles  at  the  base   of  the 
triangle  ABC :  and  it  has  also  been  proved  that  the  angle  FBC 
is  equal'to  the  anL;le  GCB,  wliich  are  the  angles  upon  the  other 
side  of  the  base.    Therefore  the  angles  at  the  base.  Sec.  Q.  E.  D. 
CoROLLARv.     Hence  every  equilateral  triangle  is  also  equi- 
angular. 


PROP.  VI.    THEOR. 


IF  two  angles  of  a  triangle  be  equal  to  one  another, 
the  sides  also  which  subtend,  or  are  opposite  tOy  the 
equal  angles  shall  be  equal  to  one  another. 


OF  EUCLID. 


19 


Let  ABC  be  a  triangle  haviiicc  the  angle  ABC  equal  to  the  Book  I. 
angle  ACB  ;  the  side  AB  is  also  equal  to  the  side  AC.  ^.— y.^^ 

For  if  AB  be  not  equal  to  AC,  one  of  them  is  greater  than 
the   other;    let  AB   be   the  greatei',   and  from  it  cut  »  off  DB  a  3.  1. 
equal  to  AC,  the  less,  and  join  DC  ;  there-  A 

fore,  because  in  the  triangles  DBC,  ACB, 
DB  is  equal  to  AC,  and  BC  common  to 
both,  the  two  sides  DB,  BC  are  equal  to 
to  the  two  AC,  CB,  each  to  each  ;  and  the 
angle  DBC  is  equal  to  the  angle  ACB; 
therefore  the  base  DC  is  equal  to  the  base 
AB,    and    the    triangle    DBC    is    equal    to 

the    triangle'' ACB,   the    less   to  the   great-         /  \\     b^.!. 

er ;  which  is  absurd.  Therefore  AB  is 
not  unequal  to  AC,  that  is,  it  is  equal  to 
it.     Wherefore,  if  two  angles,  8cc.   Q.  E.  D.     ^ 

CoR.  Hence  every  equiangular  triangle  is  also  equilateral. 


PROP.  Vn.     THEOR. 


UPON  the  same  base,  and  on  the  same  side  of  it,  See  n. 
there  cannot  be  two  triangles  that  have  their   sides 
which  are  terminated  in  one  extremity  of  the  base 
equal  to  one  another,  and  likewise  those  which  are 
terminated  in  the  other  extremity. 


If  it  be  possible,  let  there  be  two  triangles  ACB,  ADB,  up- 
on the  same  base  AB,  and  upon  the  same  side  of  it,  which  have 
their  sides  CA,  DA  terminated  in  the  extremity  A  of  the  base 
equal   to   one    another,    and  likewise  q      q 

their  sides  CB,  DB   that  are   termi- 
nated in  B. 

Join   CD  ;    then,    in    the    case    in 
which  the  vertex  of  each  of  the  tri- 
angles is  without  the  other  triangle, 
because    AC    is    equal    to    AD,    the 
angle  ACD   is   equal  *  to   the   angle 
ADC  :  but  the  angle  ACD  is  greater 
than  the  angle  BCD  ;    therefore  the 
angle  ADC  is  greater  also  than  BCD  ;     ^ 
much  more  then  is  the  angle  BDC  greater  than  the  angle  BCD. 
Again,  because  CB  is  equal  to  DB,  the  angle  BDC  is  equal  a  to  a  5. 1. 
the  angle  BCD  ;  but  it  has  been  demonstrated  to  be  greater  than 
it,  Avhich  is  impossible. 


20 


THE'  ELEMENTS 


Book  I.       Bui  ii'one  of  the  vertices,  as  D,  be  within  the  other  triangle 
*— v— '  ACB  ;  produce  AC,  AD  to  E,  F  ;  there-  E 

fore,  because  AC  is  equal  to  AD  in  the 
triangle  ACD,  the  angles  ECD,  FDC 
upon  the  other  side  of  the  base  CD  are 
a  5. 1.  equal  a  to  one  another,  but  the  angle 
ECD  is  greater  than  the  angle  BCD  ; 
wherefore  the  angle  FDC  is  likewise 
greater  than  BCD  ;  much  more  then  is 
the  angle  BDC  greater  than  the  angle 
BCD.  Again,  because  CB  is  equal  to 
DB,  the  angle  BDC  is  equal  ^  to  the  angle- 
BCD  ;  but  BDC  has  been  proved  to  be  A  B 
greater  than  the  same  BCD;  which  is  impossible.  The  case  in 
which  the  vertex  of  one  triangle  is  upon  a  side  of  the  other,  needs 
no  demonstration. 

Therefore  upon  the  same  base,  and  on  the-  same  side' of  it, 
there  cannot  be  two  triangles  that  have  their  sides  which  are 
terminated  in  one  extremity  of  the  l)ase  equal  to  one  another, 
and  likewise  those  winch  are  terminated  in  the  other  extremity* 
Q.  E.  D. 


PROP.  VIII.     THEOR. 


IF  two  triangles  have  two  sides  of  the  one  equal  to 
two  sides  of  the  other,  each  to  each,  and  have  like- 
wise their  bases  equal ;  the  angle  which  is  contained 
by  the  two  sides  of  the  one  shall  be  equal  to  the  angle 
contained  by  the  two  sides  equal  to  them,  of  the  other. 

liCt  ABC,  DEF  be  tAvw  triangles,  having  the  two  sides  AB, 
AC  equal  to  the  two  sides  DE,  DF,  each  to  each,  viz.  AB  to 
1)E,    and    AC   to    A  D     G 

DF  ;  and  also  the 
base  BC  equal  to 
the  base  EF.  The 
angle  BAC  is  e- 
qual  to  the  angle 
EDF. 

For,  if  the  tri- 
angle ABC  be  ap- 
plied to  DEI',  so        B  '       C    E  F 
that  the  point  B  be  on  E,  and  the  straight  line  BC  upon  EF : 
the  point  C  shall  also  coincide  with  the  point  F.      Because 


OF  EUCLID. 


21 


BC  is  equal  to  EF ;   therefore  BC  coinciding  with  EF,  BA  and  Book  I. 
AC  shall  coincide  with  ED  and  DF ;    for,  if  the  base  BC  coin-  y-m-ymmJ 
cides  with  the  base  EF,  but  the  sides  BA,  CA  do  not  coincide 
with  the  sides  ED,  FD,  but  have  a  diiferent  situation,  as  EG, 
FG ;  then,  upon  the  same  base  EF,  and  upon  the  same  side  of 
it,  there  can  be  two  triangles  that  have  their  sides  which  are 
terminated  in  one  extremity  of  the  base  equal  to  one  another, 
and  likewise  their  sides  terminated  in  the  other  extremity  ;   but 
this  is  impossible^;  therefore,  if  the  base  BC  coincides  with  the  a  7- 1. 
base  EF,  the  sides  BA,  AC  cannot  but  coincide  with  the  sides 
ED,  DF  ;  wherefore,  likewise,  the  angle  BAC  coincides  with  the 
angle  EDF,  and  is  equal  ^  to  it.      Therefore,  if  two  triangles,  b  8.  Ax. 
kc.     Q.  E.  D. 

PROP.  IX.     PROB. 

TO  bisect  a  given  rectilineal  angle,  that  is,  to  di- 
vide it  into  two  equal  angles. 

Let  BAC  be  the  given  rectilineal  angle  ;   it  is  required  to  bi- 
sect it. 

Take  any  point  D  in  AB,  and  from  AC  cut-off  AE  equal  to  a  3. 1. 
AD  ;  join  DE,  and  upon  it  describe  ^  A  b  1.  L 

an  equilateral  triangle  DEF  ;  then  join 
AF ;  the  straight  line  AF  bisects  the 
angle  BAC. 

Because  AD  is  equal  to  AE,  and 
AF  is  common  to  the  two  triangles 
DAF,  EAF ;  the  two  sides  DA,  AF 
are  equal  to  the  two  sides  EA,  AF, 
each  to  each ;  and  the  base  DF  is 
equal  to  the  base  EF ;  therefore  the 
angle   DAF   is   equal  ^  to  the   angle   B'  C    c  8.  1. 

EAF;  wherefore  the  given  rectilineal  angle  BAC  is  bisected  by 
the  straight  line  AF.     Which  was  to  be  done. 

PROP.  X.    PROB. 

TO  bisect  a  given  finite  straight  line,  that  is,  to 
divide  it  into  two  equal  parts. 

Let  AB  be  the  given  straight  line  ;    it  is  required  to  divide  it 
into  two  equal  parts. 

Describe  ^  upon  it  an  equilateral  triangle  ABC,  and  biscc'.  ^  a  1. 1. 
the  angle  ACB  by  the  straight  line  CD.  AB  is  cut  into  two  b  9. 1- 
equal  parts  in  the  point  D. 


22 


THE  ELEMENT 


Book  I.  Because  AC  is  equal  to  CB,  and  CD 
^■-y*^  common  to  the  two  triangles  ACD, 
BCD ;  the  two  sides  AC,  CD  are  equal 
to  BC,  CD,  each  to  each ;  and  the  angle 
ACD  is  equal  to  the  angle  BCD  ;  there- 
c  4, 1.  fore  the  base  AD  is  equal  to  the  base  ^ 
DB,  and  the  straight  line  AB  is  divided 
into  two  equal  parts  in  the  point  D. 
WJjiich  was  to  be  done. 


PROP.  XI.     FROB. 


SeeN. 


a  3,1. 
t>  1.  1. 


TO  draw  a  straight  line  at  right  angles  to  a  given 
straight  line,  from  a  given  point  in  the  same. 

Let  AB  be  a  given  straight  line,  and  C  a  point  given  in  it ;  it 
is  required  to  draw  a  straight  line  from  the  point  C  at  right  an- 
gles to  AB. 

Take  any  point  D  in  AC,  and  »  make  CE  equal  to  CD,  and 
upon  DE  describe  ''  the  equi- 
lateral triangle  DFE,  and  join 
FC  ;  the  straight  line  FC  drawn 
from  the  given  point  C  is  at 
right  angles  to  the  given  straight 
line  AB. 

Because  DC  is  equal  to  CE, 

and   FC    common    to    the    tvi^o       

?       triangles  DCF,  ECF;    the  two     A      D  C  E       B 

sides  DC,  CF  are  equal  to  the  two  EC,  CF,  each  to  each ;    and 

the  base  DF  is  equal  to  the  base  EF;    therefore  the  angle  DCF 

G  8. 1.       is   equal  <:  to  the    angle   ECF ;    and  they  are  adjacent   angles. 

But,  when  the  adjacent  angles  whi[:h  one  straight  line  m-ikes 

with  another  straight  line  are  equal  to  one  another,  each  of  them 

<110.Def.  is  called  a  right  *!  angle ;    therefore   each  of  the   angles   DCF, 

3-         ECF  is  a  right  angle.     Wherefore,  from  the  given  point  C,  in 

the  given   straigitt   line  AB,  FC  iias  been  drawn   at  I'ight  angles 

to  All.     Which  was  to  be  done. 

Cor.  By  help  of  this  ])iob!eni,  it  may  be  demonstrated,  that 
two  straight  lines  cannot  have  a  common  segment. 

If  it  be  possible,  let  the  two  straight  lines  ABC,  ABD  have 
the  segment  AB  common  to  both  of  them.  From  the  point  B 
draw  BE  at  right  angles  to- AB ;  and  because  ABC  is  a  straight 


OF  EUCLID. 


23 


line,  the  angle  CBE  is  equal  -"^  to 
the  angle  EBA  ;  in  the  same 
manner,  because  ABD  is  a 
straight  line,  the  angle  DBE  is 
equal  to  the  angle  EBA;  where- 
fore tlie  angle  DBE  is  equal  to 
the  angle  CBE,  the  less  to  the 
greater,  which  is  impossible  ; 
therefore  two  straight  lines  can- 
not have  a  common  segment. 


E 


Book  I. 


D 


A 


B 


C 


b  3-  Post. 


10.  1. 


PROP.  XII.  PROB. 

TO  draw  a  straight  line  perpendicular  to  a  given 
straight  line  of  an  unlimited  length,  from  a  given 
point  without  it. 

Let  AB  be  the  given  straight  line,  which  may  be  produced  to 
any  length  both  ways,  and  let  C  be  a  point  without  it ;   it  is  re- 
quired to  drav/  a  straight  hne  C 
perpendicular  to  AB  from  the 
point  C. 

Take  any  point  D  upon  the 
other  side  of  AB,  and  from 
the  centre  C,  at  the  distance 
CD,  describe  "^  the  circle  EGF 
meeting  AB  in  F.  G  ;  and  bi- 
sect <=  EG   in  H,  and  join  CF, 

CH,  CG  ;  the  straight  line  CH,  drawn  from  the  given  point  C, 
is  perpendicular  to  the  given  straight  line  AB. 

Because  EH  is  equal  to    HG,  and  IIC  common  to  the  two 
triangles  EHC,  GHC,  the  two  sides  EH,  HC  are  equal  to  the 
two  GH,  HC,  each  to  each  ;  and  the  base  CF  is  equal  d  to  thedl5.Def. 
base  CG  ;  therefore  the  angle  CHF  is  equal  e  to  the  angle  CHG  ;     1- 
and  they  are  adjacent  angles  ;  but  wh.?n  a  straight  line  standing  e  8.  1. 
on  a  straight  line  makes  the  adjacent  angles  equal  to  one  ano- 
ther, each  of  them  is  a  right  angle,   and  the  straight  line  which 
stands   upon   the  other  is  called  a  perpendicular  to  it ;  therefore 
from  the  given  point  C  a  perpendicular  CH  has  been  drav/n  to 
the  given  straight  line  AB.     Which  v/as  to  be  done. 

PROP.  XIII.  THEOR. 

THE  angles  which  one  straight  line  makes  with 
another  upon  the  one  side  of  it,  are  either  two  right 
angles,  or  are  together  equal  to  two  right  angles. 


24 
Book  I. 


THE  ELEMENTS 


Let  the  straight  line  AB  make  with  CD,  upon  one  side  of  it, 
the  angles  CBA,  ABD ;  these  are  either  two  right  angles,  or 
are  together  equal  tu  two  right  angles. 

For,  if  the  angle  CBx\  be  equal  to  ABD  each  of  them  is  a 
aDef.lO.  right  »  angle;   but  if  not,  from  the  point  B  draw*  BE   at  right 

A  E  A 


D 


B 


C 


D 


B 


C 


c  2.  Ax. 


b  11.  1.  angles  ^  to  CD  ;  therefore  the  angles  CBE,  EBD  are  two  right 
angles^  ;  and  because  CBE  is  equal  to  the  two  angles  CBA,  ABE 
together,  add  the  angle  EBD  to  each  of  these  equals  ;  there- 
fore the  angles  CBE,  EBD  are  equal  «=  to  the  three  angles  CBA, 
ABE,  EBD.  Again,  because  the  angle  DBA  is  equal  to  the 
two  angles  DBE,  EBA,  add  to  these  equals  the  angle  ABC  ; 
tiierefore  the  angles  DBA,  ABC  are  equal  to  the  three  angles 
DBE,  EBA,  ABC  ;  but  the  angles  CBE,  EBD  have  been  de- 
monstrated to  be  equal  to  the    same  three   angles  ;    and  things 

(1 1.  Ax.  that  are  equal  to  the  same  are  equal  ^  to  one  another ;  therefore 
the  angles  CBE,  EBD  are  equal  to, the  angles  DBA,  ABC;  but 
CBE,  EBD  are  two  right  angles ;  therefore  DBA,  ABC  are  to- 
gether equal  to  two  right  angles.  Wherefore,  when  a  straight 
line,  &c.     Q.  E.  D. 

PROP.  XIV.  THEOPv. 

IF,  at  a  point  in  a  straight  line,  two  other  straight 
lines,  upon  the  opposite  sides  of  it,  make  the  adjacent 
angles  together  equal  to  two  right  angles,  these  two 
straight  lines  shall  be  in  one  and  the  same  straight  line. 

At  the  point  B,  in  the  straight  -^ 

line  AB,  let  the  two  straight  lines 
BC,  BD,  upon  the  opposite  sides 
of  AB,  make  the  adjacent  angles 
ABC,  ABD  equal  together  to 
two  right  angles;  BD  is  in  the 
same  straight  line  with  CB. 

For,  if  BD  be  not  in  the  same' 


straight  lii^e  with  CB,  let  BE  be  C 


B 


D 


OF  EUCLID.  33 

in  the  same  straight  line  with  it;  therefore,  because  the  straight  Book  I. 

line  AB  makes  angles  with  the  straight  line  CBE,  upon  one  side  v.  ^.y—^ 

of  it,  the  angles  ABC,  ABE  are  together  equal  »  to  two  right  a  13. 1. 

angles ;  but  the  angles  ABC,  ABD  are  likewise  together  equal 

to  two  right  angles  ;   therefore  the  angles  CBA,  ABE  are  equal 

to  the  angles  CBA,  ABD :   take  away  the  common  angle  ABC, 

the  remaining  angle  ABE  is  equal  ^  to  the   remaining  angle  b  3.  Ax, 

ABD,  the  less  to  the  greater,  which  is  impossible  ;  therefore  BE 

is  not  in  the  same  straight  line  with  BC.     And,  in  like  manner, 

it  may  be  demonstrated,  that  no  other  can  be  in  the  same  straight 

line  with  it  but  BD,  which  therefore  is  in  the  same  straight  line 

with  CB.     Wherefore,  if  at  a  point,  Sec,    Q.  E.  D. 


PROP.  XV.    THEOR. 

IF  two  straight  lines  cut  one  another,  the  vertical, 

or  opposite,  angles  shall  be  equal. 

Let  the  two  straight  lines  AB,  CD  cut  one  another  in  the 
point  E ;  the  angle  AEC  shall  be  equal  to  the  angle  DEB,  and 
CEB  to  AED. 

Because  the  straight  line  AE 
makes  with  CD  the  angles  CEA, 
AED,  these  angles  are  together 

equal   »   to    two     right    angles.  ^v^ ^      *  ^^'  ^' 

Again,  because  the  straight  line 
DE  makes  with  AB  the  angles 
AED,  DEB,  these  also  are  to- 
gether equal  *  to  two  right  angles ; 
and  CEA,  AED  have  been  de- 
monstrated to  be  equal  to  two  right  angles  ;  wherefore  the  angles 
CEA,  AED  are  equal  to  the  angles  AED,  DEB.  Take  away 
the  common  angle  AED,  and  the  remaing  angle  CEA  is  equal  '^  b  3.  Ax, 
to  the  remaining  angle  DEB.  In  the  same  manner  it  can  be  de- 
monstrated that  the  angles  CEB,  AED  are  equal.  Therefore,  if 
two  straight  lines,  8cc.     Q.  E.  D. 

CoR.  1.  From  this  it  is  manifest,  that,  if  two  straight  lines  cut 
one  another,  the  angles  they  make  at  the  point  where  they  cut 
are  together  equal  to  four  right  angles. 

Cor.  2.  And,  consequently,  that  all  the  angles  made  by  any 
number  of  lines  meeting  in  one  point,  are  together  equal  to  four 
right  angles. 

D 


26 


THE  ELEMENTS 


a  10. 1. 


Book  I.  PROP.  XVI.    THEOR. 

IF  one  side  of  a  triangle  be  produced,  the  exterior 
angle  is  greater  than  either  of  the  interior  opposite 
angles. 

Let  ABC  be  a  triangle,  and  let  its  side  BC  be  produced  to  D  ; 
the  exterior  angle  ACD  is  greater  than  either  of  the  interior  op- 
posite angles  CBA,  BAG.  A 

Bisect  a  AC  in  E,  join  BE 
and  produce  it  to  F,  and 
make  EF  equal  to  BE ;  join 
also  FC,  and  produce  AC  to 
G. 

Because  AE  is  equal  to 
EC,  and  BE  to  EF ;  AE, 
EB  are  equal  to  CE,  EF, 
each  to  each  ;  and  the  angle 
AEB  is  equal  •»  to  the  angle 
CEF,  because  they  are  oppo- 
site vertical  angles;  therefore 
the  base  AB  is  equal  ^  to  the 
base     CF,     and     the     triangle 

AEB  to  the  triangle  CEF,  and  the  remaining  angles  to  the  re- 
maining angles,  each  to  each,  to  which  the  equal  sides  are  oppo- 
site ;  wherefore  the  angle  BAE  is  equal  to  the  angle  ECF;  but 
the  angle  ECD  is  greater  than  the  angle  ECF ;  therefore  the 
angle  ACD  is  greater  than  BAE :  in  the  same  manner,  if  the 
side  BC  be  bisected,  it  may  be  demonstrated  that  the  angle  BCG, 
d  15. 1.  that  is  ^,  the  angle  ACD,  is  greater  than  the  angle  ABC.  There- 
fore, if  one  side,  &c.    Q.  E.  D. 


b  15  1. 


c4. 1. 


PROP.  XVn.    THEOR. 


ANY  two  angles  of  a  triangle  are  together  less 
than  two  right  angles. 


al6. 1. 


Let  ABC  be  any  triangle  ;  any 
two  of  its  angles  together  are  less 
than  two  right  angles. 

Produce  BC  to  D ;  and  be- 
cause ACD  is  the  exterior  angle 
of  the  triangle  ABC,  ACD"  is 
greater  *  than  the  interior  and 
opposite  angle  ABC ;   to  each  of 


OF  EUCLID. 


87 


these  add  the  angle  ACB  ;  therefore  the  angles  ACD,  ACB  are  Book  L 
greater  than  the  angles  ABC,  ACB  ;    but  ACD,  ACB  are  to-  *— v— ^ 
gether  equal  ^  to  two  right  angles  ;  the-refore  the   angles  ABC,  b  ^3. 1. 
BCA  are  less  than  two  right  angles.     In  Hke  manner,  it  may  be 
demonstrated,  that  BAC,  ACB,  as  also  CAB,  ABC,  are  less  than 
two  right  angles.     Therefore,  ajiy  two  angles,  Sec.    Q.  E.  D. 


PROP.  XVIII.    THEOR. 


THE  greater  side  of  every  triangle  is  opposite  to 
2  greater  anele. 


o 

the  greater  angle. 


Let  ABC  be  a  triangle,  of  wlUeh 
the  side  AC  is  greater  than  the 
side  AB  ;  the  angle  ABC  is  also 
greater  than  the  angle  BCA. 

Because  AC  is  greater  than  AB, 
make^  AD  equal  to  AB,  and  join      /  ^ — — "■        \  a  3.  1. 

BD  ;  and  because  ADB  is  the  ex-  p 

terior  angle  of  the  triangle  BDC,  " 

it  is  greater  i»  than  the  interior  and  opposite  angle  DCB  ;  butb  16.  1. 
ADB  is  equal  <=  to  ADB,  because  the  side  AB  is  equal  to  the  side  c  5. 1. 
AD  ;  therefore  the  angle  ABD  is  likewise  greater  than  the  angle 
ACB  ;  wherefore  much  more    is  the  angle  ABC  greater  than 
ACB.     Therefore,  the  greater  side,  Sec.     Q.  E.  D. 

PROP.  XIX.    THEOR. 

THE  greater  angle  of  every  triangle  is  subtended 
by  the  greater  side,  or  bas  the  greater  side  opposite 
to  it. 

Let  ABC  be  a  triangle,  of  which  the  angle  ABC  is  greater 
than  the  angle  BCA;  the  side  AC  is  likewise  greater  than  the 
side  AB. 

For,  if  it  be  not  greater,  AC  must 
either  be  equal  to  AB,  or  less  than 
it ;  it  is  not  equal,  because  then  the 
angle  ABC  would  be  equal  »  to  the 
angle  ACB ;  but  it  is  not  ;  therefore 
AC  is  not  equal  to  AB  ;  neither  is  it 
less ;  because  then  the  angle  ABC 
would  be  lessi>  than  the  angle  ACB  ;   B  C  b  18.  1, 


a5.  1. 


28  THE  ELEMENTS 

Book  I.  but  it  is  not  ;  therefore  the  side  AC  is  not  less  than  AB  ;  and  it 

•— "v^^  has  been  shown  that  it  is   not  equal  to  AB ;  therefore  AC  is 

greater  than  AB.     Wherefore,  the  greater  angle,  See.    Q.  E.  D. 


PROP.  XX.  THEOR. 

See  N.  ANY  two  sidcs  of  a  triangle  are  together  greater 
than  the  third  side. 

Let  ABC  be  a  triangle  ;  any  two  sides  of  it  together  are 
greater  than  the  third  side,  viz.  the  sides  BA,  AC  greater  than 
the  side  BC  ;  and  AB,  BC  greater  than  AC  ;  and  BC,  CA  great- 
er than  AB. 

Produce  BA  to  the  point  D,   and 

a  3.  1.     make  ^  AD  equal  to  AC  ;  and  join 
DC. 

Because  DA  is  equal  to  AC,  the 

b  5.  1.     angle   ADC  is  likewise   equal  ''  to 
ACD  ;  but  the  angle  BCD  is  great- 
er than  the  angle  ACD  ;    therefore 
the  angle  BCD  is  greater  than  the     B 
angle  ADC  ;    and  because  the  angle  BCD  of  the  triangle  DCB 

c  19.  1.  is  greater  than  its  angle  BDC,  and  that  the  greater  <=  side  is  op- 
posite to  the  greater  angle  ;  therefore  the  side  DB  is  greater  than 
the  side  BC  ;  but  DB  is  equal  to  BA  and  AC  ;  therefore  the  sides 
BA,  AC  are  greater  than  BC.  In  the  same  manner  it  may  be 
demonstrated,  that  the  sides  AB,  BC  are  greater  than  CA,  and  BC, 
CA  greater  than  AB.     Therefore,  any  two  sides,  Sec.     Q.  E.  D. 


PROP.  XXL  THEOR. 

See  N.  IF,  from  the  ends  of  the  side  of  a  triangle,  there  be 
drawn  two  straight  lines  to  a  point  within  the  triangle, 
these  shall  be  less  than  the  other  two  sides  of  the  tri- 
angle, but  shall  contain  a  greater  angle. 

Let  the  two  straight  lines  BD,  CD  be  drawn  from  B,  C,  the 
ends  of  the  side  BC  of  the  triangle  ABC,  to  the  point  D  within 
it  ;  BD  and  DC  are  less  than  the  other  two  sides  BA,  AC 
of  the  triangle,  but  contain  an  angle  BDC  greater  than  the  an- 
gle BAC. 

Produce  BD  to  E  ;  and  because  two  sides  of  a  triangle  are 
greater  than  the  third  side,  the  two  sides  BA,  AE  of  the  tri- 


OF  EUCLID. 


29 


angle  ABE  are  greater  than  BE.     To  each  of  these  add  EC ;  Book  I. 
therefore  the  sides  BA,  AC   are  A  ^'--r-^ 

greater  than  BE,  EC :  again, 
because  the  two  sides  CE,  ED 
of  the  triangle  CED  are  great- 
er than  CD,  add  DB  to  each  of 
these ;  therefore  the  sides  CE, 
EB  are  greater  than  CD,  DB ; 
but  it  has  been  shown  that  BA, 
AC  are  greater  than  BE,  EC ; 
xnuch  more  then  are  BA,  AC 
greater  than  BD,  DC. 

Again,  because  the  exterior  angle  of  a  triangle  is  greater  than 
the  interior  and  opposite  angle,  the  exterior  angle  BDC  of  the 
triangle  CDE  is  greater  than  CED ;  for  the  same  reason,  the 
exterior  angle  CEB  of  the  triangle  ABE  is  greater  than  BAC; 
and  it  has  been  demonstrated  that  the  angle  BDC  is  greater  than 
the  angle  CEB ;  much  more  then  is  the  angle  BDC  greater  than 
the  angle  BAC.     Therefore,  if  from  the  ends  of,  &c.     Q.  E.  D* 


PROP.  XXII.     PROB. 


TO  make  a  triangle  of  which  the  sides  shall  be  See  n. 
equal  to  three  given  straight  lines,  but  any  two  what- 
ever of  these  must  be  greater  than  the  third  ^  a20.  i 


Let  A,  B,  C  be  the  three  given  straight  lines,  of  which  any 
two  whatever  are  greater  than  the  third,  viz.  A  and  B  greater 
than  C  ;  A  and  C  greater  than  B  ;  and  B  and  C  than  A.  It  is 
required  to  make  a  triangle  of  which  the  sides  shall  be  equal  to 
A,  B,  C,  each  to  each. 

Take  a  straight  line  DE  terminated  at  the  point  D,  but  un- 
limited towards  E,  and 
make  ^  DF  equal  to  A, 
FG  to  B,  and  GH  equal 
to  C ;  and  from  the  centre 
F,  at  the  distance  FD,  de- 
scribe ''  the  circle  DKL ; 
and  from  the  centre  G,  at 
the  distance  GH,  describe 
^  another  circle  HLK ; 
and  join  KF,  KG;  the 
triangle  KFG  has  its  sides 
equal  to  the  three  straight  lines  A,  B,  C. 

Because  the  point  F  is  the  centre  of  the  circle  DKL,  FD 


a  3.1. 


.Post. 


30 


THE  ELEMENTS 


Book  I.  equal  ^  to  FK  ;  but  FD  is  equal  to  the  straight  line  A  ;  there- 
^— >r— '  fore  FK  is  equal  to  A :  again,  because  G  is  the  centre  of  the 
cl5.Def.  circle  LKH,  GH  is  equal  «  to  GK  ;  but  GH  is  equal  to  C; 
therefore  also  GK  is  equal  to  C ;  and  FG  is  equal  to  B ;  there- 
fore the  three  straight  lines  KF,  FG,  GK  are  equal  to  the  three 
A,  B,  C :  and  therefore  the  triangle  KFG  has  its  three  sides 
KF,  FG,  GK  equal  to  the  three  given  straight  lines  A,  B,  C. 
Which  was  to  be  done. 


a  22. 1. 


b  8.  1. 


PROP.  XXIII.     PROB. 

AT  a  given  point  in  a  given  straight  line,  to  make 
a  rectilineal  angle  equal  to  a  given  rectilineal  angle. 

Let  AB  be  the  given  straight  line,  and  A  the  given  point  in 
it,  and  DCE  the  given  rectilineal  angle :  it  is  required  to  make 
an  angle  at  the  given 
point  A  in  the  given 
straight  line  AB,  that 
shall  be  equal  to  the 
given  rectilineal  an- 
gle DCE. 

Take  in  CD,  CE 
any  points  D,  E,  and 
join  DE  ;  and  make  ^ 
the  triangle  AFG,  the 
sides  of  vi'hich  shall 
be  equal  to  the  three 
straight  lines  CD,  DE,  CE,  so  that  CD  be  equal  to  AF,  CE  to 
AG,  and  DE  to  FG,  and  because  DC,  CE  are  equal  to  FA, 
AG,  each  to  each,  and  the  base  DE  to  the  base  FG  ;  the  angle 
DCE  is  equal  ^  to  the  angle  FAG.  Therefore,  at  the  given 
point  A,  in  the  given  straight  line  AB,  the  angle  FAG  is  made 
equal  to  the  given  rectilineal  angle  DCE.  Which  was  to  be 
done. 


PROP.  XXIV.     THEOR. 

See  N.  IF  two  tnanglcs  have  two  sides  of  the  one  equal  to 
two  sides  of  the^other,  each  to  each,  but  the  angle 
contained  by  the  two  sides  of  one  of  them  greater 
than  the  angle  contained  by  the  two  sides  equal  to 
them,  of  the  other ;  the  base  of  that  which  has  the 
greater  angle  shall  be  greater  than  the  base  of  the  other. 


OF  EUCLID. 


SI 


Let  ABC,  DEF  be  two  triangles  which  have  the  two  sides  Book  I, 
AB,  AC  equal  to  the  two  DE,  DF,  each  to  each,  viz.  AB  equal  '— v— ' 
to  DE,  and  AC  to  DF  ;  but  the  angle  BAC  greater  than  the  an- 
gle EDF ;  the  base  BC  is  also  greater  than  the  base  EF. 

Of  the  two  sides  DE,  DF,  let  DE  be  the  side  which  is  not 
greater  than  the  other,  and  at  the  point  D,  in  the  straight  line 
DE,  make  »  the  angle  EDG  equal  to  the  angle  BAC  ;   and  make  a  23. 1. 
DG  equal  b  to  AC  or  DF,  and  join  EG,  GF.  b  3. 1. 

Because  AB  is  equal  to  DE,  and  AC  to  DG,  the  two  sides 
BA,  AC  are  equal  to  the  two  ED,  DG,  each  to  each,  and  the 
angle  BAC  is  equal         A  D 

to  the  angle  EDG ; 
therefore  the  base  BC 
is  equal  =  to  the  base 
EG;  and  because  DG 
is  equal  to  DF,  the 
angle  DFG  is  equal"^ 
to  the  angle  DGF; 
but  the  angle  DGF  is 
greater  than  the  an- 
gle EGF  ;  therefore 
the  angle  DFG  is  greater  than  EGF ;  and  much  more  is  the  angle 
EFG  greater  than  the  angle  EGF ;  and  because  the  angle  EFG 
of  the  triangle  EFG  is  greater  than  its  angle  EGF,  and  that 
the  greater  e  side  is  opposite  to  the  greater  angle;  the  side  EG^  ^^' ^' 
is  therefore  greater  than  the  side  EF;  but  EG  is  equal  to  BC ; 
and  therefore  also  BC  is  greater  than  EF.  Therefore,  if  two 
triangles,  Sec.    Q.  E.  D. 


c4. 1. 


d5.  1. 


PROP.  XXV.    THEOR. 


IF  two  triangles  have  two  sides  of  the  one  equal  to 
two  sides  of  the  other,  each  to  each,  but  the  base  of 
the  one  greater  than  the  base  of  the  other ;  the  angle 
also  contained  by  the  sides  of  that  which  has  the 
greater  base,  shall  be  greater  than  the  angle  contained 
by  the  sides  equal  to  them,  of  the  other. 

Let  ABC,  DEF  be  two  triangles  which  have  the  two  sides 
AB,  AC  equal  to  the  two  sides  DE,  DF,  each  to  each,  viz.  AB 
equal  to  DE,  and  AC  to  DF ;  but  the  base  CB  is  greater  than 
the  base  EB' ;  the  angle  BAC  is  likewise  greater  than  the  anele 
EDF. 


32 


THE  ELEMENTS 


Book  I.       For,  if  it  be  not  greater,  it  must  either  be  equal  to  it,  or  less; 
^— v"-~^  but  the  angle  BAG  is  not  equal  to  the  angle  EDF,  because  then 

the  base  BC  would 
a4fl-      beequalatoEF;  but  A  D 

it  is  not ;    therefore 

the  angle  BAG  is  not 

equal  to   the    angle 

EDF;   neither  is  it 

less ;    because   then 

the  base  BG   would 
b  24. 1.    be    less  ^  than    the 

base  EF ;   but  it  is 

not ;    therefore    the 

angle  BAG  is  not  less  than  the  angle  EDF ;  and  it  was  shown 

that  it  is  not  equal  to  it ;  therefore  the  angle  BAG  is  greater 

than  the  angle  EDF.   Wherefore,  if  two  triangles,  &c.   Q.  E.  D. 


PROP.  XXVI.     THEOR. 

IF  two  triangles  have  two  angles  of  one  equal  to 
two  angles  of  the  other,  each  to  each ;  and  one  side 
equal  to  one  side,  viz.  either  the  sides  adjacent  to 
the  equal  angles,  or  the  sides  opposite  to  equal  angles 
in  each ;  then  shall  the  other  sides  be  equal,  each  to 
each; 
angle  of  the  other. 


and  also  the  third  angle  of  the  one  to  the  third 


Let  ABG,  DEF  be  two  triangles  which  have  the  angles  ABG, 
BGA  equal  to  the  angles  DEF,  EFD,  viz.  ABG  to  DEF,  and 
BGA  to  EFD  ;  also  one  side  equal  to  one  side ;  and  first  let 
those  sides  be  equal  which  are  adjacent  to  the  angles  that  are 
equal  in  the  two  tri-     A  D 

angles,  viz.  BG  to 
EF  ;  the  other  sides  q. 
shall  be  equal,  each 
to  each,  viz.  AB  to 
DE,  and  AG  to  DF  ; 
and  the  third  angle 
BAG  to  the  third  an- 
gle EDF. 

For,  if  AB  be  not      B  G         E  F 

equal  to  DE,  one  of  them  must  be  the  greater.  Let  AB  be  the 
greater  of  the  two,  and  make  BG  equal  to  DE,  and  join  GC  ; 
therefore,  because  BG  is  equal  to  DE,  and  BG  to  EF,  the  two 


OF  EUCLID.  3S 

sides  GB,  BC  are  equal  to  the  two  DE,  EF,  each  to  each ;   and  Book  I. 
the  angle  GBC  is  equal  to  the  angle  DEF ;    therefore  the  base  ^■— v— > 
GC  is  equal  ^  to  the  base  DF,  and  the  triangle  GBC  to  the  tri-  a  4. 1. 
angle  DEF,  and  the  other  angles  to  the  other  angles,  each  to 
each,  to  which  the  equal  sides  are  opposite ;  therefore  the  angle 
GCB  is  equal  to  the  angle  DFE ;    but  DFE  is,  by  the  hypothe- 
sis, equal  to  the  angle  BCA ;   wherefore  also  the  angle  BCG  is 
equal  to  the  angle  BCA,  the  less  to  the  greater,  which  is  im- 
possible ;  therefore  AB  is  not  unequal  to  DE,  that  is,  it  is  equal 
to  it ;    and  BC  is  equal  to  EF ;    therefore  the  two  AB,  BC  are 
equal  to  the  two  DE,  EF,  each  to  each ;    and  the  angle  ABC  is 
equal  to  the  angle  DEF ;    the  base  therefore  AC  is  equal  »  to 
the  base  DF,  and  the  third  angle  BAG  to  the  third  angle  EDF. 

Next,    let    the    sides 
which   are   opposite  to  A  D 

equal  angles  in  each 
triangle  be  equal  to  one 
another,  viz.  AB  to 
DE ;  likewise  in  this 
case,  the  other  sides 
shall  be  equal,  AC  to 
DF,  and  BC  to  EF :  and 
also  the  third  angle 
BAG  to  the  third  EDF. 

For,  if  BC  be  not  equal  to  EF,  let  BC  be  the  greater  of  them, 
and  make  BH  equal  to  EF,  and  join  AH ;  and  because  BH  is 
equal  to  EF,  and  AB  to  DE ;  the  two  AB,  BH  are  equal  to 
the  two  DE,  EF,  each  to  each  ;  and  they  contain  equal  angles ; 
therefore  the  base  AH  is  equal  to  the  base  DF,  and  the  triangle 
ABH  to  the  triangle  DEF,  and  the  other  angles  shall  be  equal, 
each  to  each,  to  which  the  equal  sides  are  opposite ;  therefore 
the  angle  BHA  is  equal  to  the  angle  EFD ;  but  EFD  is  equal 
to  the  angle  BCA ;  therefore  also  the  angle  BHA  is  equal  to 
the  angle  BCA,  that  is,  the  exterior  angle  BHA  of  the  triangle 
AHC  is  equal  to  its  interior  and  opposite  angle  BCA;  which  is 
impossible'*;  wherefore  BC  is  not  unequal  to  EF,  that  is,  it  isbl6. 1. 
equal  to  it;  and  AB  is  equal  to  DE ;  therefore  the  two  AB, 
BC  are  equal  to  the  two  DE,  EF,  each  to  each ;  and  they  con- 
tain equal  angles ;  wherefore  the  base  AC  is  equal  to  the  base 
DF,  and  the  third  angle  BAG  to  the  third  angle  EDF.  There- 
fore, if  two  triangles,  &c.    Q.  E.  D. 


34  THE  ELEMENTS 

Book  I.  PROP.  XXVII.    THEOR. 

IF  a  straight  line  falling  upon  two  other  straight 
lines  makes  the  alternate  angles  equal  to  one  another, 
these  two  straight  lines  shall  be  parallel. . 

Let  the  straight  line  EF,  which  falls  upon  the  two  straight 
lines  AB,  CD,  make  the  alternate  angles  AEF,  EFD  equal  to 
one  another ;  AB  is  parallel  to  CD. 

For,  if  it  be  not  parallel,  AB  and  CD  being  produced  shall 
meet  either  towards  B,  D,  or  towards  A,  C ;  let  them  be  pro- 
duced and  meet  towards  B,  D,  in  the  point  G ;  therefore  GEF 
a  16. 1.  is  a  triangle,  and  its  exterior  angle  AEF  is  greater  » than  the  in- 
terior and   opposite    angle 


EFG  ;    but  it  is  also  equal  ^    / 


to  it,  which  is  impossible;     . ^f  ^ 

therefore  AB  and  CD  be- 


ing produced  do  not  meet 
towards  B,  D.  In  like  man- 
ner it  may  be  demonstrat- 
ed, that  they  do  not  meet 
towards   A,  C ;    but  those 

straight  lines  which  meet  neither  way,  though  produced  ever  s» 
b35.Def.  far,  are  parallel  ^  to  one  another.    AB  therefore  is^parallel  to  CD. 
Wherefore,  if  a  straight  line,  Sec     Q.  E.  D. 

PROP.  XXVIII.    THEOR. 

IF  a  straight  line  falling  upon  two  other  straight 
lines  makes  the  exterior  angle  equal  to  the  interior 
and  opposite  upon  the  same  side  of  the  line  ;  or  makes 
the  interior  angles  upon  the  same  side  together  equal 
to  two  right  angles ;  the  two  straight  lines  shall  be 
parallel  to  one  another. 

Let  the  straight  line  EF,  which  E 

,falls  upon  the  two  straight  lines 
AB,  CD,  make  the  exterior  angle 
EGB  equal  to  the  interior  and  op-  A  ■ 
posite  angle  GHD  upon  the  same 
side  ;  or  make  the  interior  angles 
on  the  same  side  BGH,  GHD  toge- 
ther equal  to  two  right  angles  ;  AB 
is  parallel  to  CD. 

13ecause  the  angle  EGB  is  equal 
to  the  angle  GHD,  and  the  angle 


OF  EUCLID.  35 

EGB   equal  *  to   the  angle  AGH,    the  angle  AGH  is  equal  Book  I. 
to  the  angle  GHD  ;  and  they  are  the  alternate  angles ;  therefore  *>-— v— -^ 
AB  is  parallel  ^  to  CD.    Again,  because  the  angles  BGH,  GHD  a  15.  1. 
are  equal  <=  to  two  right  angles ;  and  that  AGH,  BGH  are  also  b  27. 1, 
equal  d  to  two  i^ight  angles  ;  the  angles  AGH,  BGH  are  equal  cBy  hyp. 
to  the   angles  BGH,    GHD:    take   away  the   common  angle d  13. 1 
BGH ;  therefore  the  remaining  angle  AGH  is  equal  to  the  re- 
maining angle  GHD  ;  and  they  are  alternate  angles  ;  therefore 
AB  is   parallel   to   CD.     Wherefore,    if  a   straight  line,   Sec. 
Q.  E.  D. 

PROP.  XXIX.     THEOR. 


IF  a  straight  line  fall  upon  two  parallel  straight  See  the 
lines,  it  makes  the  alternate  angles  equal  to  one  ano-  "iJJs"°" 
ther  ;   and  the  exterior  angle  equal  to  the  interior  and  position 
opposite  upon  the  same  side  ;    and  likewise  the  two 
interior  angles  upon  the  same  side  together  equal  to 
two  right  angles. 

Let  the  straight  line  EF  fall  upon  the  parallel  straight  lines 
AB,  CD ;  the  alternate  angles,  AGH,  GHD  are  equal  to  one 
another;  and  the  exterior  angle  EGB  is  equal  to  the  interior 
and  opposite,  upon  the  same  side  E 

GHD ;  and  the  two  interior  an- 
gles BGH,  GHD  upon  the  same  . . 
side  are  together  equal  to  two 
right  angles. 

For,  if  AGH  be  not  equal  to  ^ 
GHD,  one  of  them  must  be  great- 
er than  the  other  ;   let  AGH  be 
the  greater  ;  and  because  the  an- 
gle AGH  is  greater  than  the  an- 
gle GHD,  add  to  each  of  them  the  angle  BGH  ;  therefore  the 
angles  AGH,  BGH  are  greater  than  the  angles  BGH,  GHD  ;  but 
the  angles  AGH,  BGH  are  equal  »  to  two  right  angles ;  therefore  a  13. 1. 
the  angles  BGH,  GHD  are  less  than  two  right  angles  ;  but  those 
straight  lines  which,  with  another  straight  line  falling  upon  them, 
make  the  interior  angles  on  the  same  side  less  than  two  right 
angles,  do  meet*  together  if  continually  produced  ;  therefore  »  12.  ax. 
the  straight  lines  AB,  CD,  if  produced  far  enough,  shall  meet  ;  See  the 
but  they  never  meet,  since  they  are  parallel  by  the  hypothesis  ;  "o'es  on 
therefore  the  angle  AGH  is  not  unequal  to  the  angle  GHD,  that  *^'^.  P'"' 
is,  it  is  equal  to  it ;   but  the  angle  AGH  is  equal  ^   to  the  angle  n'*'7' 
EGB;   therefore  likewise  EGB  is  equal  to  GHD  ;   add  to  each 


36  THE  ELEMENTS 

Book  I.  of  these  the  angle  BGH  ;  therefore  the  angles  EGB,  BGH  are 

^— V— '  equal  to  the  angles  BGH,  GHD  ;   but  EGB,  BGH  are  equal  c~ 

c  13.  1.     to  two   right  angles ;   therefore   also  BGH,  GHD  are  equal  to 

two  right  angles.     Wherefore,  if  a  straight  line,  Sec.    Q.  E.  D. 


PROP.  XXX.     THEOR. 

STRAIGHT  lines  which  are  parallel  to  the  same 
straight  line  are  parallel  to  one  another. 

Let  AB,  CD  be  each  of  them  parallel  to  EF  ;  AB  is  also  pa- 
rallel to  CD. 

Let  the  straight  line  GHK  cut  AB,  EF,  CD ;   and  because 
GHK    cuts    the    parallel    straight 
lines  AB,  EF,  the  angle  AGH  is 

a  29. 1.  equal  ^  to  the  angle  GHF.  Again, 
because  the  straight  line  GK 
cuts  the  parallel  straight  lines 
EF,  CD,  the  an^le  GHF  is  equal 
to  the  angle  GKD  ^  ;  and  it  was 
shown  that  tlie  angle  AGK  is 
equal  to  the  angle  GHF  ;  there- 
fore also  AGK  is  equal  to  GKD  ; 
and    they    are    alternate    angles  ; 

b  27. 1.    therefore  AB  is  parallel  ^  to  CD.    Wherefore,  straight  lines,  &c. 
Q.  E.  D. 


PROP.  XXXL    PROB. 

TO  draw  a  straight  line  through  a  given  point  pa- 
rallel to  a  given  straight  line. 

Let  A  be  the  given  point,  and  BC  the  given  straight  line  ;  it 
is  required  to  draw  a  straight  line 

through  the  point  A,  parallel  to  the  E  A  F 

straight  line  BC. 

In  BC  take  any  point  D,  and  join 
AD  ;   and  at   the  point   A,   in   the 

A  23. 1.     straight  line  AD,  make  *  the  angle 

DAE  equal  to  the  angle  ADC  ;  and  B  D  C 

produce  the  straight  line  EA  to  F. 

Because  the  straight  line   AD,  which  meets  the  two  straight 

lines  BC,  EF,  makes  the  alternate  angles  EAD,  ADC  equal  to 

b  27. 1.    one  another,  EF  is  parallel  ^  to  BC     Therefore  the  straight  line 


OF  EUCLID. 


37 


EAF  is  drawn  through  the  given  point  A  parallel  to  the  given  Book  I. 
straisrht  line  BC.     Which  was  to  be  done.  V-i-Y^i^* 


PROP.  XXXII.     THEOR, 


IF  a  side  of  any  triangle  be  produced,  the  exterior 
angle  is  equal  to  the  two  interior  and  opposite  angles  ; 
and  the  three  interior  angles  of  every  triangle  are  equal 
to  two  right  angles. 

Let  ABC  be  a  triangle,  and  let  one  of  its  sides  BC  be  produ- 
ced to  D  ;  the  exterior  angle  ACD  is  equal  to  the  two  interior 
and  opposite  angles  CAB,  ABC ;  and  the  three  interior  angles 
of  the  triangle,  viz.  ABC,  BCA,  CAB,  are  together  equal  to  two 
right  angles. 

Through  the   point   C  A 

draw  CE  parallel  »  to  the  y\  ^E  a  31. 1. 

straight  line  AB  ;  and  be- 
cause AB  is  parallel  to 
CE,  and  AC  meets  them, 
the  alternate  angles  BAC, 

ACE  are  equal  b.   Again,  B  v^  ^      b  29. 1. 

because  AB  is  parallel  to  CE,  and  BD  falls  upon  them,  the  ex- 
terior angle  ECD  is  equal  to  the  interior  and  opposite  angle 
ABC  ;  but  the  angle  ACE  was  shown  to  be  equal  to  the  angle 
BAC  ;  therefore  the  whole  exterior  angle  ACD  is  equal  to  the 
two  interior  and  opposite  angles  CAB,  ABC  ;  to  these  equals 
add  the  angle  ACB,  and  the  angles  ACD,  ACB  are  equal  to  the 
three  angles  CBA,  BAC,  ACB  ;  but  the  angles  ACD,  ACB  are 
equal  c  to  two  right  angles :  therefore  also  the  angles  CB  A,  c  13.  1. 
BAC,  ACB  are  equal  to  two  right  angles.  Wherefore,  if  a  side 
of  a  triangle,  Sec.     Q.  E.  D. 

Cor.  1.  AH  the  interior  angles 
of  any  rectilineal  figure,  together 
with  four  right  angles,  are  equal 
to  twice  as  many  right  angles  as 
the  figure  has  sides. 

For  any  rectilineal  figure 
ABCDE  can  be  divided  into  as 
many  triangles  as  the  figure  has 
sides,  by  drawing  straight  lines 
from  a  point  F  within  the  figure 
to  each   of  its   angles.     And,   by 


38 


THE  ELEMENTS 


Book  I.  ^^^  preceding  proposition,  all  the  angles  of  these  triangles  are 

\,^.^^mmJ  equal  to  twice  as  many  right  anglts  as  there  are  triangles,  that 

is,  as  there  are  sides  of  the  figure  ;  and  the  same  angles  are  equal 

to  the  angles  of  the  figure,  together  with  the  angles  at  the  point 

a  2.  Cor.  F,  which  is  the  common  vertex  of  the  triangles  ;  that  is  ^,  to- 

15. 1.       gether  with  four  right  angles.     Therefore  all  the  angles  of  the 

figure,  together   with   four  right  angles,  are  equal  to  twice  as 

many  right  angles  as  the  figure  has  sides. 

CoR.  2,    All  the  exterior  angles  of  any  rectilineal  figure,  are 
together  equal  to  four  right  angles. 

Because  every  interior  angle 
ABC,  with  its  adjacent  exterior 
I>13.  1.  ABD,  is  equal  •>  to  two  right 
angles  ;  therefore  all  the  interior, 
together  with  all  the  exterior 
angles  of  the  figure,  are  equal  to 
twice  as  many  right  angles  as 
there  are  sides  of  the  figure  :  D 
that  is,  by  the  foregoing  corol- 
lary, they  are  equal  to  all  the 
interior  angles  of  the  figure,  to- 
gether with  four  right  angles  ;  therefore  all  the  exteripr  an- 
gles are  equal  to  four  right  angles. 


PROP.  XXXIII.    THEOR. 

THE  Straight  lines  wliich  join  the  extremities  of 
two  equal  and  parallel  straight  lines,  towards  the 
same  parts,  are  also  themselves  equal  and  parallel. 


Let  AB,  CD  be  equal  and  pa- 
rallel straight  lines,  and  joined  to- 
wards the  same  parts  by  the  straight 
lines  AC,  BD  ;  AC,  BD  arc  also 
equal  and  parallel. 

Join  BC ;  and  because  AB  is 
parallel    to    CD,     and    IIC     meets  C  D 

91  29. 1.  them,  the  alternate  angles  ABC,  BCD  are  equal  a  ;  and  because 
AB  is  equal  to  CD,  and  BC  common  to  the  two  triangles  ABC, 
DCB,  the  two  sides  AB,  BC  are  equal  to  the  two  DC,  CB  ; 
and  the  angle  ABC  is  equal  to  the  angle  BCD  ;  therefore  the 

b  4  1  base  AC  is  equal  •»  to  the  base  BD,  and  ihe  triangle  ABC  to  the 
triangle  BCD,  and  the  other  angles  to  the  other  angles  ^^  each 
to  each,  to  which  the  equal  sides  are  opposite  :  therefore  th? 


OF  EUCLID.  3S 

angle  ACB  is  equal  to  the  angle  CBD  ;  and  because  the  straight  Book  I. 
line  BC  meets  the  two  straight  lines  AC,  BD,  and  makes  the  al-  ^■— y— ' 
ternate  angles  ACB,  CBD  equal  to  one  another,  AC  is  parallel 
^  to  BD ;  and  it  was  shown  to  be  equal  to  it.     Therefore,  straight  c  27. 1. 
lines,  &c.     Q.  £.  D. 


PROP.  XXXIV.    THEOR. 

THE  opposite  sides  and  angles  of  parallelograms 
are  equal  to  one  another,  and  the  diameter  bisects 
them,  that  is,  divides  them  into  two  equal  parts. 

N.  B.  A  parallelogram  is  a  four  sided  figure^  of 
which  the  opposite  sides  are  parallel ;  and  the  diameter 
is  the.  straight  line  joining  tvjo  of  its  opposite  angles. 


Let  ACDB  be  a  parallelogram,  of  which  BC1s  a  diameter; 
the  opposite  sides  and  angles  of  the  figure  are  equal  to  one  an- 
other ;  and  the  diameter  BC  bisects  it. 

Because  AB  is  parallel  to  CD,     A B 

and   BC   meets   them,    the   alter- 
nate angles  ABC,  BCD  are  equal 

=*   to    one    another ;    and   because        \  y^  \        a  29. 

AC   is    parallel   to   BD,    and   BC 
meets  them,  the  alternate  angles 

ACB,   CBD    are   equal  ^   to    one  C  D 

another;  wherefore  the  two  triangles  ABC,  CBD  have  two 
angles  ABC,  BCA  in  one,  equal  to  two  angles  BCD,  CBD  in 
the  other,  each  to  each,  and  one  side  BC  common  to  the  two 
triangles,  which  is  adjacent  to  their  equal  angles ;  therefore 
their  other  sides  shall  be  equal,  each  to  each,  and  the  third 
angle  of  the  one  to  the  third  angle  of  the  other  ^,  viz.  the  b  25. 1. 
side  AB  to  the  side  CD,  and  AC  to  BD,  and  the  angle  BAC 
equal  to  the  angle  BDC ;  and  beciuse  the  angle  ABC  is  equal 
to  the  angle  BCD,  and  the  angle  CBD  to  the  angle  ACB,  the 
whole  angle  ABD  is  equal  to  the  whole  angle  ACD :  and  the 
angle  BAC  has  been  shown  to  be  equal  to  the  angle  BDC ; 
therefore  the  opposite  sides  and  angles  of  parallelograms  are 
equal  to  one  another ;  also,  their  diameter  bisects  them  ;  for  AB 
being  equal  to  CD,  and  BC  common,  the  two  AB,  BC  are 
equal  to  the  two  DC,  CB,  each  to  each ;  and  the  angle  ABC  is 


40 


THE  ELEMENTS 


Book  I.  equal  to  the  angle  BCD ;  therefore  the  triangle  ABC  is  equal 
*'''""'  <=  to  the  triangle  BCD,  and  the  diameter  BC  divides  the  parallelo- 
c  4. 1.      gram  ACDB  into  two  equal  parts.     Q.  E.  D. 


PROP.  XXXV.    THEOR. 


SeeN. 


PARALLELOGRAMS  upon  the  same  base,  and 
between  the  same  parallels,  are  equal  to  one  another. 


figures. 


a  34.  1. 


See  the^        Let  the  parallelograms  ABCD,  EBCF  be  upon  the  same  base 
-^^J^'^^'^BC,  and  between  the  same  parallels  AF,  BC  ;  the  parallelogram 
ABCD  shall  be  equal  to  the  parallelogram  EBCF. 

If  the  sides  AD,  DF  of  the  paral- 
lelograms ABCD,  DBCF  opposite  to 
the  base  BC  be  terminated  in  the  same 
point  D,  it  is  plain  that  each  of  the 
parallelograms  is  double  ^  of  the  trian- 
gle BDC  ;  and  they  are  therefore  equal 
to  one  another. 

But,  if  the  sides  AD,  EF,  opposite 
to  the  base  BC  of  the  parallelograms 

ABCD,  EBCF,  be  not  terminated  in  the  same  point,  then,  be- 
cause ABCD  is  a  parallelogram,  AD  is  equal  ^  to  BC  ;  for  the 
same  reason,  EF  is  equal  to  BC ;  wherefore  AD  is  equal  ^  to 
EF,  and  DE  is  common  ;  therefore  the  whole,  or  the  remain- 
c  2.  or  3.  der  AE,  is  equal  <=  to  the  whole,  or  the  remainder  DF  ;  AB  also 
Ax.         is  equal  to  DC ;  and  the  two  EA,  AB  are  therefore  equal  to  the 


b  1.  A> 


d  19. 1. 

4.  1. 


Ax. 


two  FD,  DC,  each  to  each  ;  and  the  exterior  angle  FDC  is 
equal  ^  to  the  interior  EAB ;  therefore  the  base  EB  is  equal  to 
the  base  FC,  and  the  triangle  EAB  equal  <=  to  the  triangle  FDC ; 
take  the  triangle  FDC  from  the  trapezium  ABCF,  and  from  the 
same  trapezium  take  the  triangle  EAB;  the  remainders  there- 
fore are  equal  f,  that  is,  the  parallelogram  ABCD  is  equal  to  the 
parallelogram  EBCF.  Therefore,  parallelograms  upon  the  same 
base,  &c.    -Q.  E.  D.  ^ 


OF  EUCLID. 


PROP.  XXXVI.    THEOR. 


PARALLELOGRAMS  upon   equal  bases,  and 
between  the  same  parallels,  are  equal  to  one  another. 

Let  ABCD,  EFGH  be 
parallelograms  upon  equal 
bases  BC,  FG,  and  be- 
tween the  same  parallels 
AH,  BG ;  the  parallelo- 
gram ABCD  is  equal  to 
EFGH. 

Join  BE,  CH;  and  be-    B  C  F  G 

cause  BC  is  equal  to  FG,  and  FG  to  »  EH,  BC  is  equal  to  EH  ;  a  34  1, 
and  they  are  parallels,  and  joined  towards  the  same  parts  by  the 
straight  lines  BE,  CH :  but  straight  lines  which  join  equal  and 
parallel  straight  lines  towards  the  same   parts,  are  themselves 
equal  and  parallel  ^  ;  therefore  EB,  CH  are  both  equal  and  pa-  b  33.  1- 
rallel,  and  EBCH  is  a  parallelogram  ;  and  it  is  equal  ^  to  ABCD,  c  35. 1. 
because  it  is  upon  the  same  base  BC,  and  between  the  same 
parallels  BC,  AD  :  for  the  like  reason,  the  parallelogram  EFGH 
is  equal  to  the  same  EBCH:  therefore  also  the  parallelogram 
ABCD  is  equal  to  EFGH.     Wherefore,   parallelograms,   &c. 
Q.  E.  D. 


PROP.  XXXVII.    THEOR. 

TRIANGLES  upon  the  same  base,  and  between 
the  same  parallels,  are  equal  to  one  another. 

Let  the  triangles  ABC,  DBC  be  upon  the  same  base  BC  and 
between    the   same   parallels   ^^  An  v 

AD,  BC  :  the  triangle  ABC     ^  ^     " 

is  equal  to  the  triangle  DBC. 

Produce  AD  both  ways  to 
the  points  E,  F,  and  through 
B  draw  »  BE. parallel  to  CA  ; 
and  through  C  draw  CF  pa- 
rallel to  BD  :  therefore  each 
of  the  figures  EBCA,  DBCF 
is  a  parallelogram ;  and  EBCA  is  equal  ^  to  DBCF,  because  b  35. 1. 
they  are  upon  the  same  base  BC,  and  between  the  same  parallels 
BC,   EF  ;   and  the  triangle  ABC  is  the  half  of  the  parallelo- 

F 


31. 1 


42  THE  ELEMENTS 

Book  I.  gram  EBCA,  because  the  diameter  AB  bisects  «  it ;  and  the  tri- 
angle DBC  is  the  half  of  the  parallelogram  DBCF,  because  the 
diameter  DC  bisects  it:  but  the  halves  of  equal  things  are 
equal  d;  therefore  the  triangle  ABC  is  equal  to  the  triangle 
DBC.     Wherefore,  triangles,  Sec     Q.  E.  D. 


PROP.  XXXVIIL    THEOR. 

TRIANGLES  upon  equal  bases,  and  between  the 
same  parallels,  are  equal  to  one  another. 

Let  the  triangles  ABC,  DEF  be  vjpon  equal  bases  BC,  EF,  and 
between  the  same  parallels  BF,  AD  :  the  triangle  ABC  is  equal 
to  the  triangle  DEF. 

Produce  AD  both  ways  to  the  points  G,  H,  and  through  B 

a  SI.  1-  draw  BG  parallel  »  to  CA,  and  through  F  draw  FH  parallel 
to  ED :  then  each  of     p  .  p.  tt 

the   figures    GBCA,     ^  A  u  ti 

DEFH  i5  a  parallel- 
ogram,     and      they 

b  36. 1.  are  equal  ^  to  one 
another,because  they 
are  upon  equal  bases 
BC,  EF,  and  be- 
tween the  same  pa- 

c34. 1.  rallels  BF,  GH  ;  and  the  triangle  ABC  is  the  half  <=  of  the  pa- 
rallelogram GBCA,  because  the  diameter  AB  bisects  it ;  and  the 
triangle  DEF  is  the  half  ^  of  the  parallelogram  DEFH,  because 
the  diameter  DF  bisects  it :  but  the  halves  of  equal  things  are 

d  7.  Ax.  equal  ^ ;  therefore  the  triangle  ABC  is  equal  to  the  triangle 
DEF.     Whprefore,  triangles,  8cc.     Q,  E.  D. 


PROP.  XXXIX.    THEOR. 

EQUAL  triangles  upon  the  same  base,  and  upon 
the  same  side  of  it,  are  between  the  same  parallels. 

Let  the  equal  triangles  ABC,  DBC  be  upon  the  same  base 
BC,  and  upon  the  same  side  of  it ;  they  are  between  the  same 
parallels. 

Join  AD  ;  AD  is  parallel  to  BC  ;  for,  if  it  is  not,  through  the 
»  SI.  1-    point  A  draw  ^  AE  parallel  to  BC,  and  join  EC :  the  triangle 


OF  EUCLID. 


43 


ABC  is  equal  •»  to  the  triangle  EBC,  because  it  is  upon  the  same  Book  I. 
base  BC,  and  between  the  same  paral- 
lels BC,  AE:  but  the  triangle  ABC  is 
equal  to  the  triangle  BDC  ;  therefore 
also  the  triangle  BDC  is  equal  to  the 
triangle  EBC,  the  greater  to  the  less, 
which  is  impossible:  therefore  AE  is 
not  parallel  to  BC.  In  the  same  man- 
ner, it  can  be  demonstrated  that  no  other 
line  but  AD  is  parallel  to  BC  ;  AD  is 
therefore  parallel  to  it. 
Q.  E.  D. 


B  C 

Wherefore,    equal    triangles,    &c. 


PROP.  XL.    THEOR. 

EQUAL  triangles  upon  equal  bases,  in  the  same 
straight  line,  and  towards  the  same  parts,  are  between 
the  same  parallels. 

Let  the  equal  triangles  ABC,  DEF  be  upon  equal  bases  BC, 
EF,    in    the    same    straight 
line    BF,    and    towards    the 
same    parts ;    they    are  be- 
tween the  same  parallels. 

Join  AD ;  AD  is  paral- 
lel to  BC  :  for,  if  it  is  not, 
through  A  draw  ^  AG  pa- 
rallel to  BF,  and  join  GF :  „  r  v  t? 
the  triangle  ABC  is  equal  »>  ^  *-  £.  ^  b  38. 1, 
to  the  triangle  GEF,  because  they  are  upon  equal  bases  BC,  EF, 
and  between  the  same  parallels  BF,  AG  :  but  the  triangle  ABC 
is  equal  to  the  triangle  DEF  ;  therefore  also  the  triangle  DEF 
is  equal  to  the  triangle  GEF,  the  greater  to  the  less,  which  is 
impossible  :  therefore  AG  is  not  parallel  to  BF  :  and  in  the  same 
manner  it  can  be  demonstrated  that  there  is  no  other  parallel  to 
it  but  AD ;  AD  is  therefore  parallel  to  BF.  Wherefore,  equal 
triangles,  &c.    Q.  E.  D. 


a  31. 1. 


PROP.  XLI.    THEOR. 


IF  a  parallelogram  and  triangle  be  upon  the  same 
base,  and  between  the  same  parallels ;  the  parallelo- 
gram shall  be  double  of  the  triangle. 


44 


THE  ELEMENTS 


Book  I.       Let  the  parallelogram  ABCD  and  the  triangle  EBC  be 

«— V— '  the  same  base  BC,  and  between  the  same  parallels  BC,  AE 
parallelogram  ABCD   is  double  of  the 
triangle  EBC. 

Join    AC ;    then   the   triangle   ABC 

a  37. 1.  is  equal  *  to  the  triangle  EBC,  because 
they  are  upon  the  same  base  BC,  and 
between   the  same  parallels   BC,   AE. 

b  34.  1.  But  the  parallelogram  ABCD  is  double ^ 
of  the  triangle  ABC,  because  the  diame- 
ter AC  divides  it  into  two  equal  parts ; 
wherefore  ABCD  is  also  double  of  the 
triangle  EBC.  Therefore,  if  a  parallelo- 
gram, Sec.    Q.  E.  D. 


upon 
;the 


PROP.  XLIL    PROB. 


A    F 


TO  describe  a  parallelogram  that  shall  be  equal  to 
a  given  triangle,  and  have  one  of  its  angles  equal  to 
a  given  rectilineal  angle. 

Let  ABC  be  the  given  triangle,  and  D  the  given  rectilineal 
angle.  It  is  required  to  describe  a  parallelogram  that  shall  be 
equal  to  the  given  triangle  ABC,  and  have  one  of  its  angles 
equal  to  D. 

a  10. 1.         Bisect  a  BC  in  E,  join  AE,  and  at  the  point  E  in  the  straight 

b  23. 1.    line  EC  make  ^  the  angle  CEF  equal  to  D  ;  and  through  A  draw 

c  31. 1.  ^  AG  parallel  to  EC,  and  through 
C  draw  CG  «=  parallel  to  EF : 
therefore  FECG  is  a  parallelo- 
gram :  and  because  BE  is  equal 
to  EC,  the  triangle  ABE  is  like- 

d  38.  1.  wise  equal  ^  to  the  triangle  AEC, 
since  they  are  upon  equal  bases 
BE,  EC,  and  between  the  same 
parallels  BC,  AG  :  therefore  the 
triangle  ABC  is  double  of  the 
triangle    AEC:     and    the    paral- 

e4l.  1.  lelogram  FECG  is  likewise  double  e  of  the  triangle  AEC,  be- 
cause it  is  upon  the  same  base,  and  belv/een  the  same  parallels  : 
therefore  the  parallelogram  FECG  is  equal  to  the  triangle 
ABC,  and  it  has  one  of  its  angles  CEF  equal  to  the  given 
angle  D.     Wherefore  there  has  been  described  a  parallelogram 


OF  EUCLID.  45 

FECG  equal  to  a  given  triangle  ABC,  having  one  of  its  angles  Book  I. 
CEF  equal  to  the  given  angle  D.     Which  was  to  be  done.  ^— v-*' 


PROP.  XLIII.    THEOR. 

THE  complements  of  the  parallelograms  which 
are  about  the  diameter  of  any  parallelogram,  are 
equal  to  one  another. 

Let  ABCD  be  a  parallelogram,  of  which  the  diameter  is  AC 
and  EH,  FG  the  parallelograms 
about  AC,  that  is,  through  which 
AC  passes,  and  BK,  KD  the 
other  parallelograms  which 
make  up  the  whole  figure  ABCD, 
which  aTe  therefore  called  the 
complements ;  the  complement 
BK  is  equal  to  the  complement 
KD. 

Because  ABCD  is  a  paral- 
lelogram, and  AC  its  diame- 
ter, the  triangle  ABC  is  equal  *  to  the  triangle  ADC  :  and  a  34<  !♦ 
because  EKHA  is  a  parallelogram,  the  diameter  of  which  is 
AK,  the  triangle  AEK  is  equal  to  the  triangle  AHK :  by  the 
same  reason,  the  triangle  KGC  is  equal  to  the  triangle  KFC : 
then,  because  the  triangle  AEK  is  equal  to  the  triangle  AHK, 
and  the  triangle  KGC  to  KFC;  the  triangle  AEK  together 
with  the  triangle  KGC  is  equal  to  the  triangle  AHK  together 
with  the  triangle  KFC  :  but  the  whole  triangle  ABC  is  equal 
to  the  whole  ADC ;  therefore  the  remaining  complement  BK 
is  equal  to  the  remaining  complement  KD.  Wherefore,  the 
complements,  &c.    Q.  E.  D. 


PROP.  XLIV.     PROB. 

TO  a  given  straight  line  to  apply  a  parallelogram, 
which  shall  be  equal  to  a  given  triangle,  and  have 
one  of  its  angles  equal  to  a  given  rectilineal  angle. 

Let  AB  be  the  given  straight  line,  and  C  the  given  triangle, 
and  D  the  given  rectilineal  angle.  It  is  required  to  apply  to 
the  straight  line  AB  a  parallelogram  equal  to  the  triangle  C, 
and  having  an  angle  equal  to  D. 


46 


THE  ELEMENTS 


Book  I.       Make  »  the 

^"--^-^^  parallelogr«im 

a  42.1,  BEFG  equal 
to  the  triangle 
C,  and  having 
the  angle  EBG 
equal  to  the 
angje  D,  so 
that  BE  be 
in  thC'  same 
straight      line 

b  31. 1.  with  AB,  and  produce  EG  to  H ;  and  through  A  draw  •'  AH  pa- 
rallel to  BG  or  EF,  and  join  HB.  Then,  because  the  straight 
line  HF  falls  upon  the  parallels  AH,  EF,  the  angles  AHF,  HFE 

c  29. 1.  are  together  equal *=  to  two  right  angles;  wherefore  the  angles 
BHF,  HFE  are  less  than  two  right  angles  :  but  straight  lines 
which  with  another  straight  line  make  the  interior  angles  upon 

d  12.  Ax.  the  same  side  less  than  two  right  angles  do  meet  •*,  if  produced 
far  enough  :  therefore  HB,  FE  shall  meet,  if  produced  ;  let  them 
meet  in  K,  and  through  K  draw  KL  parallel  to  EA  or  FH,  and 
produce  HA,  GB  to  the  points  L,  M :  then  HLKF  is  a  paral- 
lelogram, of  which  the  diameter  is  HK,  and  AG,  ME  are  the 
parallelograms  about  HK ;  and  LB,  BF  are  the  complements  ; 
therefore  LB  is  equal  e  to  BF ;  but  BF  is  equal  to  the  triangle 
C ;  wherefore  LB  is  equal  to  the  triangle  C :  and  because  the 
angle  GBE  is  equal  f  to  the  angle  ABM,  and  likewise  to  the 
angle  D  ;  the  angle  ABM  is  equal  to  the  angle  D  :  therefore 
the  parallelogram  LB  is  applied  to  the  straight  line  AB,  is  equal 
to  the  triangle  C,  and  has  the  angle  ABM  equal  to  the  angle  D. 
Which  was  to  be  done. 


e  43.  i: 


f  15. 1. 


PROP.  XLV.     PROB. 


TO  describe  a  parallelogram  equal  to  a  given  rec- 
tilineal figure,  and  having  an  angle  equal  to  a  given 
rectilineal  angle. 


a42. 1. 


b44. 1. 


Let  ABCD  be  the  given  rectilineal  figure,  and  E  the  given 
rectilineal  angle.  It  is  required  to  describe  a  parallelogram  equal 
to  ABCD,  and  having  an  angle  equal  to  E. 

Join  DB,  and  describe  »  the  parallelogram  FH  equal  to  the 
triangle  ADB,  and  having  the  angle  HKF  equal  to  the  angle  E; 
and  to  the  straight  line  GH  apply  ^  the  parallelogram  GM  equal 


OF  EUCLID.  47 

to  the  triangle  DBC,  having  the  angle  GHM  equal  to  the  angle  Book  I. 
E  ;  and  because  the  angle  E  is  equal  to  each  of  the  angles  FKH,  *— v— ' 
GHM,  the  angle  FKH  is  equal  to  GHM  ;  add  to  each  of  these 
the  angle  KHG  ;  therefore  the  angles  FKH,  KHG  are  equal  to 
the   angles   KHG, 

GHM ;  but  FKH,    A  ^  F  G       L 

KHG    are     equal 

c  to  two  right  an-        \  /  \      I  t         /  /         /  c  29. 1. 

gles  ;  therefore 
also  KHG,  GHM 
are  equal  to  two 
right  angles  ;  and 
because  at  the 
point  H  in  the 
straight  line  GH 
the  two  straight  lines  KH,  H!M,  upon  the  opposite  sides  of  it, 
make  the  adjacent  angles  equal  to  two  right  angles,  KH  is  in 
the  same  straight  line  ^  with  HM  ;  and  because  the  straight  line  d  14. 1. 
HG  meets  the  parallels  KM,  FG,  the  alternate  angles  MHG, 
HGF  are  equal  «  :  add  to  each  of  these  the  angle  HGL  :  there- 
fore the  angles  MHG,  HGL  are  equal  to  the  angles  HGF, 
HGL  :  but  the  angles  MHG,  HGL  are  equal  <=  to  two  right  an- 
gles ;  wherefore  also  the  angles  HGF,  HGL  are  equal  to  two 
right  angles,  and  FG  is  therefore  in  the  same  straight  line  with 
GL :  and  because  KF  is  parallel  to  HG,  and  HG  to  ML  ;  KF  is 
parallel*  to  ML  :  and  KM,  FL  are  parallels  ;  wherefore  KFLM  e  30. 1. 
is  a  parallelogram  ;  and  because  the  triangle  ABD  is  equal  to 
the  parallelogram  HF,  and  the  triangle  DBC  to  the  parallelo- 
gram GM  ;  the  whole  rectilineal  figure  ABCD  is  equal  to  the 
whole  parallelogram  KFLM  ;  therefore  the  parallelogram  KFLM 
has  been  described  equal  to  the  given  rectilineal  figure  ABCD 
having  the  angle  FKM  equal  to  the  given  angle  E.  Which  was 
to  be  done. 

Cor.  From  this  it  is  manifest  how  to  a  given  straight  line  to 
apply  a  parallelogram,  which  shall  have  an  angle  equal  to  a  given 
rectilineal  angle,  and  shall  be  equal  to  a  given  rectilineal  figure, 
viz.  by  applying  ^  to  the  given  straight  line  a  parallelogram  b  44. 1. 
equal  to  the  first  triangle  ABD,  and  having  an  angle  equal  to  the 
given  angle. 


*8  THE  ELEMENTS 

Book  I. 


PROP.  XLVI.    PROB. 


TO  describe  a  square  upon  a  given  straight  line. 

Let  AB  be  the  given  straight  line ;  it  is  required  to  describe  a 

square  upon  AB. 
a  11. 1.        From  the  point  A  draw  a  AC  ^t  right  angles  to  AB  ;  and 
b  3. 1.      make  i"  AD  equal  to  AB,  and  through  the  point  D  draw   DE 
c  31. 1.     parallel  ^  to  AB,  and  through  B  draw  BE  parallel  to  AD  ;  there- 
d34. 1.    fore  ADEB  is  a  parallelogram:  whence  AB  is  equal d  to  DE, 

and   AD  to  BE  :   but  BA  is  equal  to      C 

AD  ;  therefore  the  four  straight  lines 

BA,  AD,  DE,    EB   are   equal   to  one 

another,  and  the  parallelogram  ADiiB 

is  equilateral,    likewise  all   its  angles 

are  right  angles ;  because  the  straight 

line    AD   meeting   the    parallels    AB, 

DE,  the  angles  BAD,  ADE  are  equal 
e  29. 1.     *=  to  two  right  angles ;  but  BAD  is  a 

right  angle;   therefore  also  ADE  is  a 

right  angle  ;  but  the  opposite   angles 

of  parallelograms  are  equal  ^  ;  tliere-      A 

fore  each  of  the  opposite  angles  ABE, 

BED  is  a  right  angle  ;  wherefore  the  figure  ADEB  is  rectan- 
gular, and  it  has  been  demonstrated  that  it  is  equilateral  ;  it  is 

therefore  a  square,  and  it  is  described  upon  the  given  straight 

line  AB.     Which  was  to  be  done. 

CoR.    Hence   every   parallelogram  that  has  one  right  angle 

has  all  its  angles  right  angles. 


B 


PROP.  XLVn.    THEOR. 


IN  any  right  angled  triangle,  the  square  which  is 
described  upon  the  side  subtending  the  right  angle 
is  equal  to  the  squares  described  upon  the  sides 
which  contain  the  right  angle. 

Let  ABC  be  a  right  angled  triangle,  having  the  right  angle 
BAC  ;  the  square  described  upon  the  side  BC  is  equal  to  the 
squares  described  upon  BA,  AC. 
^46.1.        On  BC  describe  »  the  square  BDEC,   and  on  BA,  AC  the 


OF  EUCLID. 


49 


i  14. 1. 


squares  GB,  HC  ;  and  through  A  draw  ^  AL  parallel  to  BD  or  Book  1. 

CE,  and  join  AD,  FC  ;  then,  because  each  of  the  angles  BAG.  ^-— y««> 

BAG  is   a   right   angle  <=,    the  G  b  31. 1. 

two    straight    lines    AC,  AG,  X\  ».  «30.Def. 

cpon    the    opposite    sides    of 

AB,  make  with  it  at  the  point 

A  the  adjacent    angles  equal  F 

to  two  right  angles  ;    therefore 

CA    is    in    the    same  straight 

line  ^  Avith  AG  ;  for  the  same 

reason,    AB  and    AH  are    in 

the    same    straight  line  ;    and 

because  the  angle  DBC  is    e- 

qual   to  the   angle  FBA,  each 

of  them   being    a  right  angle, 

add   to    each  the  angle    ABC, 

and   the  whole   angle  DBA  is 

equal  ^  lo  the  whole  FBC  ;  and  e  2,  Ax. 

because  the  two  sides  AB,  BD  are  equal  to  the  two  FB,  BC, 

each    to   each,   and  the   angle   DBA  equal  to  the  angle  FBC  ; 

therefore  the  base  AD  is  equal  f  to  the  base  FC,   and  the  trian-  f  4.  1. 

gle  ABD    to  the  triangle  FBC  :  now  the  parallelogram  BL  is 

double  s  of  the  triangle  ABD,  because  they  are  upon  the  same  g  41. 1. 

base  BD,    and  between  the  same  parallels,  BD,  AL  ;    and  the 

square  GB  is  double  of  the  triangle  FBC,  because  these  also  are 

upon  the  same  base  FB,  and  between  the  same   parallels  FB, 

GC.     But   the    doubles   of  eqvials  are  equal  ^  to  one  another :  h  6.  Ax. 

therefore  the  parallelogram  BL  is  equal  to  the  square  GB :  and 

in  the  same  manner,  by  joining  AE,  BK,  it  is  demonstrated  that 

the  parallelogram  CL  is  equal  to  the  square  HC  :  therefore  the 

whole  square  BDEC  is  equal  to  the  two  squares  GB,  HC  ;  and 

the  square  BDEC  is  described  upon  the  straight   line  BC,  and 

the  squares  GB,  HC  upon  BA,  AC  :  wherefore  the  square  upon 

the  side  BC  is  equal  to  the  squares  upon  the  sides  BA,  AC« 

Therefore,  in  any  right  angled  triangle,  &c.     Q.  E.  D. 


PROP.  XLVin.    THEOR. 


IF  the  square  described  upon  one  of  the  sides  of 
a  triangle  be  equal  to  the  squares  described  upon  the 
other  two  sides  of  it ;  the  angle  contained  by  these 
two  sides  is  a  right  angle. 


G 


5.0 


THE  ELEMENTS,  £cc. 


Book  I.       If  the  square  described  upon  BC,  one  of  the  sides  of  the  tw- 
>-- v*-^  angle  ABC,  be  equal  to  the  squares  upon  the  other  sides  BA, 

AC,  the  angle  BAC  is  a  right  angle, 
a  11. 1.        From  the  point  A  draw »  AD  at  right  angles  to  AC,  and 

make  AD  equal   to  BA,  and  join  DC  :  then,  because  DA  is 

equal  to  AB,  the  square  of  DA  is  equal  to  D 

the  square  of  AB :    to  each  of  these  add 

the  square  of  AC  ;    therefore  the   squares 

of  DA,  AC   are  equal   to   the  squares  of 
b  47. 1.     BA,  AC  :  but  the  square  of  DC  is  equal  ^ 

to  the  squares  of  DA,  AC,  because  DAC 

is  a  right  angle  ;  and  the  square  of  BC,  by 

hypothesis,  is  equal  to  the  squares  of  BA, 

AC  ;  therefore  the  square  of  DC  is  equal 

to   the   square  of  BC  ;  and  therefore  also     ^  C 

the  side  DC  is  equal  to  the  side  BC.     And  because  the  side  DA 

is  equal  to  AB,  and  AC  common  to  the  two  triangles  DAC, 

BAC,  the  two  DA,  AC  are  equal  to  the  two  BA,  AC  ;  and  the 

base  DC  is  equal  to  the  base  BC  ;  therefore  the  angle  DAC  is 
c  8.  1.      equal  c  to  the  angle  BAC :  but  DAC  is  a  right  angle  ;  therefore 

also  BAC  is  a  right  angle.  Therefore,  if  the  square,  &c»  Q.  E.  D. 


THE 


ELEMENTS  OF  EUCLID. 


BOOK  IL 


DEFINITIONS. 


I. 

Every  right  angled  parallelogram  is  said  to  be  contained  by  Book  II. 

any  two  of  the  straight  lines  which  contain  one  of  the  right  Vi^v"^ 

angles. 

II. 
In  every  parallelogram,  any  of  the  parallelograms  about  a  dia- 
meter,   together     with      the  P 

two   complements,   is   called      A , , .  D 

a  gnomon.      '  Thus  the  pa- 

'  rallelogram    HG,   together 

'  with  the  complements  AF, 
'    '  FC,  is  the  gnomon,  which 

*  is  more  briefly  expressed     H  -~y^ — jj^ 

'  by    the   letters    AGK,    or 

'  EHC,    which    are    at  the      B 

*  opposite  angles  of  the  pa- 
'  rallelograms  v/hich  make  the  gnomon.' 


F 


PROP.  I.     THEOR. 


IF  there  be  two  straight  lines,  one  of  which  is 
divided  into  any  number  of  parts ;  the  rectangle  con- 
tained  by  the  two  straight  lines,  is  equal  to  the  rect- 
angles contained  by  the  undivided  line  and  the 
several  parts  of  the  divided  line. 


52 


THE  ELEMENTS 


Book  II.      Let  A  and  BC  be  two  straight  lines  ;  and  let  BC  be  divided 

*— V— ^  into  any  parts  in  the  points  D,  E  ;  the  rectangle  contained  by 
the  straight  lines  A,  BC  is  equal 
to  the  rectangle  contained  by  A, 
BD,  together  with  that  contained 
by  A,  DE,  and  that  contained  by 
A,  EC. 

a  11.1.         From  the   point  B    draw  ^    BF 

at  right  angles  to  BC,    and  make  G 

b  3. 1.      BG  equal''    to    A;    and    through 

c31. 1.     G    drawc     GH    parallel    to    BC ; 

and  through  D,  E,  C  draw  ^  DK,  F 

EL,    CH    parallel     to    BG  ;     then 

the  rectangle  BH  is  equal  to  the  i'ectangles  BK,  DL,  EH  ;  and 

BH  is  contained  by  A,  BC,  for  it  is  contained  by  GB,  BC,  and 

GB  is  equal  to  A  ;   and  BK  is  contained  by  A,  BD,   for  it  is 

contained  by  GB.  BD,  of  which  GB  is  equal  to  A  ;  and  DL  is 

d  34. 1.  contained  by  A,  DE,  because  DK,  that  is  'i,  BG,  is  equal  to  A  ; 
and  in  like  manner  the  rectangle  EH  is  contained  by  A,  EC  : 
therefore  the  rectangle  contained  by  A,  BC  is  equal  to  the  se- 
veral rectangles  contained  by  A,  BD,  and  by  A,  DE ;  and  also 
by  A,  EC.  Wherefore,  if  there  be  two  straight  lines,  Sec. 
Q.  E.  D. 


PROP.   n.     THEOR. 


IF  a  straight  line  be  divided  into  any  two  parts, 
the  rectangles  contained  by  the  whole  and  each  of 
the  parts  are  together  equal  to  the  square  of  the 
whole  line. 


•i.  46.  1. 
bSl.l. 


Let  the  straight  line  AB  be  divided  into 
any  two  pans  in  the  point  C  ;  the  rect- 
angle contained  by  AB,  BC,  together  with 
the  rectangle  *  AB,  AC,  shall  be  equal  to 
the  square  of  AB. 

Upon  AB  describe  »  the  square  ADEB, 
and  through  C  draw  ^  CF,  parallel  to  AD 
or  BE  ;  then  AE  is  equal  to  the  rectangles 
AF,  CE  ;  and  AK  is  the  square  of  AB  ; 
and  AF  is  the  rectangle  contained  by  BA, 


*  N.  B.  To  avoid  repeating  the  word  contained  too  frequently,  the  rectangle 
contaiiipcl  by  tv/o  straiglit  lines  AB,  AC  is  sometimes  simply  called  the  rect- 
angle AB,  AC. 


OF  EUCLID.  5«5 

AC  ;  for  it  is  contained  by  DA,  AC,  of  which  AD  is  equal  to  Book  II. 
AB  ;  and  CE  is  contained  by  AB,  BC,  for  BE  is  equal  to  AB;  ^— •v^»"' 
therefore  the  rectangle  contained  by  AB,  AC,  together  Avith  the 
rectangle  AB,  BC,  is  equal  to  the  square  of  AB.   If  therefore  a 
straight  line,  &c.     Q.  E.  D. 


PROP.  III.     THEOR. 

IF  a  straight  line  be  divided  into  any  two  parts, 
the  rectangle  contained  by  the  whole  and  one  of  the 
parts  is  equal  to  the  rectangle  contained  by  the  two 
parts,  together  with  the  square  of  the  foresaid  part. 

Let  the  straight  line  AB  be  divided  into  any  two  parts  in  the 
point  C  ;  the  rectangle  AB,  BC  is  equal  to  the  rectangle  AC, 
CB,  together  with  the  square  of  BC. 

Upon  BC  describe  ^  the  square 
CDEB,  and  produce  ED  to  F,  and 
through  A  draw  •»  AF  parallel  to 
CD  or  BE  ;  then  the  rectangle  AE 
is  equal  to  the  rectangles  AD,  CE  : 
and  AE  is  the  rectangle  contained 
by  AB,  BC,  for  it  is  contained  by 
AB,  BE,  of  which  BE  is  equal  to 
BC  ;  and  AD  is  contained  by  AC, 
CB,  for  CD  is  equal  to  CB  ;  and  DB 
is  the  square  of  BC  ;  therefore  the  rectangle  AB,  BC  is  equal  to 
the  rectangle  AC,  CB,  together  with  the  square  of  BC.  If  there- 
fore a  straight  line,  See.    Q.  E.  D. 


B  a  46. 1. 


b  31. 1. 


PROP.  IV.     THEOR. 

IF  a  Straight  line  be  divided  into  any  two  parts, 
the  square  of  the  whole  line  is  equal  to  the  squares 
of  the  two  parts,  together  with  twice  the  rectangle 
contained  by  the  parts. 

Let  the  straight  line  AB  be  divided  into  any  two  parts  in  C  ; 
the  square  of  AB  is  equal  to  the  squares  of  AC,  CB,  and  to 
twice  the  rectangle  contained  by  AC,  CB. 


54 


THE  ELEMENTS 


Book  II.  Upon  AB  describe  *  the  square  ADEB,  and  join  BD,  and 
through  C  draw  ^  CGF  parallel  to  AD  or  BE,  and  through  G 
draw  HK  parallel  to  AB  or  DE  :  and  because  CF  is  parallel  to 
AD,  and  BD  falls  upon  them,  the  exterior  angle  BGC  is  equal 
c  to  the  interior  and  opposite  angle  ADB ;  but  ADB  is  equal 
^  to  the  angle  ABD,  because  BA  is  equal  to  AD,  being  sides  of 
a  square ;    wherefore    the  angle   CGB       A  C  B 

is  equal  to  the  angle  GBC  ;  and  there- 
fore the  side  BC  is  equal  ^  to  the  side 
CG :  but  CB  is  equal  *^  also  to  GK, 
and  CG  to  BK ;  wherefore  the  figure  H 
CGKB  is  equilateral :  it  is  likewise 
rectangular  ;  for  CG  is  parallel  to  BK, 
and  CB  meets  them  ;  the  angles  KBC, 
GCB  are  therefore  equal  to  two  right 
an.s:les  ;   and    KBC   is    a   right   angle ; 


e  6.1. 
£34.1 


G  ^ 


K 


E 


D  F 

wherefore  GCB  is  a  right  angle ;  and  therefore  also  the  angles 
f  CGK,  GKB,  opposite  to  these,  are  right  angles,  and  CGKB  is 
rectangular:  but  it  is  also  equilateral,  as  was  demonstrated; 
wherefore  it  is  a  square,  and  it  is  upon  the  side  CB :  for  the 
same  reason  HF  also  is  a  square,  and  it  is  upon  the  side  HG, 
which  is  equal  to  AC :  therefore  HF,  CK  are  the  squares  of 
g  43. 1.  AC,  CB ;  and  because  the  complement  AG  is  equal  s  to  the 
complement  GE,  and  that  AG  is  the  rectangle  contained  by 
AC,  CB,  for  GC  is  equal  to  CB  ;  therefore  GE  is  also  equal 
to  the  rectangle  AC,  CB  :  wherefore  AG,  GE  are  equal  to 
twice  the  rectangle  AC,  CB  :  and  HF,  CK  are  the  squares  of 
AC,  CB  ;  wherefore  the  four  figures  HF,  CK,  AG,  GE  are 
equal  to  the  sqiiares  of  AC,  CB,  and  to  twice  the  rectangle 
AC,  CB  ;  but  HF,  CK,  AG,  GE  make  up  the  whole  figure 
ADEB,  which  is  the  square  of  AB  :  therefore  the  square  of  AB 
is  equal  40  the  squares  of  AC,  CB,  and  twice  the  rectangle  ACj 
CB.     Wherefore,  if  a  straight  line,  &c.     Q.  E.  D. 

Cor.   From  the  demonstration,  it  is  manifest,  that  the  paral- 
lelograms about  the  diameter  of  a  square  are  likewise  squares. 


OF  EUCLID. 


PROP.  V.    THEOR. 


IF  a  straight  line  be  divided  into  two  equal  parts, 
and  also  into  two  unequal  parts;  the  rectangle  con- 
tained by  the  unequal  parts,  together  with  the  square 
of  the  line  between  the  points  of  section,  is  equal  to 
the  square  of  half  the  line. 

Let  the  straight  line  AB  be  divided  into  two  equal  parts  in 
the  point  C,  and  into  two  unequal  parts  at  the  point  D  ;  the 
rectangle  AD,  DB,  together  with  the  square  of  CD,  is  equal  to 
the  square  of  CB. 

Upon  CB  describe  »  the  square  CEFB,  join  BE,  and  through  ^  46. 1. 
D  draw  b  DHG  parallel  to  CE  or  BF  ;   and  through  H  draw  b  31. 1. 
KLM  parallel  to  CB  or  EF  ;  and  also  through  A  draw  AK  pa-    - 
rallel   to  CL   or  BM  :    and   because   the   complement   CH   is 
equal  =  to  the   complement   HF,  to   each  of  these   add   DM ;  ^  43. 1. 
therefore     the     whole     CM 
is  equal  to   the  whole  DF  ;         A  C  D      B 

but   CM  is  equal  ^   to   AL,         ( i      71      d  36. 1. 

because  AC  is  equal  to 
CB ;  therefore  also  AL  is 
equal  to  DF.  To  each  of 
these  add  CH,  and  the 
whole  AH  is  equal  to  DF 
and  CH :  but  AH  is  the 
rectangle  contained   by  AD, 

DB,  for  DH  is  equal  e  to  DB  ;  and  DF  together  with  CH  is  the  eCor,4.2. 
gnomon  CMG  ;  therefore  the  gnomon  CMG  is  equal  to  the 
rectangle  AD,  DB  :  to  each  of  these  add  LG,  which  is  equal  « 
to  the  square  of  CD  ;  therefore  the  gnomon  CMG,  together 
with  LG,  is  equal  to  the  rectangle  AD,  DB,  together  with  the 
square  of  CD:  but  the  gnomon  CMG  and  LG  make  up  the 
whole  figure  CEFB,  which  is  the  square  of  CB  :  therefore  the 
rectangle  AD,  DB,  together  with  the  square  of  CD,  is  equal  to 
the  square  of  CB.  Wherefore,  if  a  straight  line,  Sec.  Q.  E.  D. 
From  this  proposition  it  is  manifest,  that  the  difference  of  the 
squares  of  two  unequal  lines  AC,  CD,  is  equal  to  the  rectangle 
contained  by  their  sum  and  difference. 


THE  ELEMENTS 


PROP.  VI.    THEOR. 


a  46.  1. 
bSl.  1. 


IF  a  straight  line  be  bisected,  and  produced  to  any 
point,  the  rectangle  contained  by  the  whole  line  thus 
produced,  and  the  part  of  it  produced,  together  with 
the  square  of  half  the  line  bisected,  is  equal  to  the 
square  of  the  straight  line  which  is  made  up  of  the 
half  and  the  part  produced. 

Let  the  straight  line  AB  be  bisected  in  C,  and  produced  to 
the  point  D  ;  the  rectangle  AD,  DB,  together  with  the  square 
of  CB,  is  equal  to  the  square  of  CD. 

Upon  CD  describe  »  the  square  CEFD,  join  DE,  and  through 
B  draw  b  BHG  parallel  to  CE  or  DF,  and  through  H  draw  KLM 
parallel  to  AD  or  EF,  and  also  through  A  draw  AK  parallel  toCL 


B 


D 


L 

K 

/ 

/ 

M 


G        F 


or  DM  :  and  because  AC   is 
equal  to  CB,    the   rectangle 

c  36.  1.     AL  is  equal   <=■  to  CH  ;   but 

d  43.  1.  CH  is  equal  ^  to  HF  ;  there- 
fore also  AL  is  equal  to 
HF  :  to  each  of  these  add 
CM ;  therefore  the  whole 
AM  is  equal  to  the  gnomon 
CMC  :  and  AM  is  the  rect- 
angle contained  by  AD,  DB, 

eCor-4.2.  for  DM  is  equal  ^  to  DB  :  therefore  the  gnomon  CMG  is  equal 
to  the  rectangle  AD,  DB  :  add  to  each  of  these  LG,  which  is 
equal  to  the  square  of  CB  ;  therefore  the  rectangle  AD,  DB,  to- 
gether Avith  the  square  of  CB,  is  equal  to  the  gnomon  CMG 
and  the  figure  LG  :  but  the  gnomon  CMG  and  LG  make  up  the 
whole  figure  CEFD,  which  is  the  square  of  CD  ;  therefore 
the  rectangle  AD,  DB,  together  with  the  square  of  CB,  is 
equal  to  the  square  of  CD.  Wherefore,  if  a  straight  line,  &c. 
Q.  E.  D. 

PROP.  Vn.     THEOR. 

IF  a  straight  line  be  divided  into  any  two  parts,  the 
squares  of  the  whole  line,  and  of  one  of  the  parts,  are 
equal  to  twice  the  rectangle  contained  by  the  whole 
and  that  part,  together  with  the  square  of  the  other 
part. 


Let   the    straight  line  AB  be  divided  into  any  two   parts  in 


OF  EUCLID. 


57 


the  ijoint  C  ;  the  squares  of  AB,  BC  are  equal  to  twice  the  rect-Book  II. 
angle  AB,  BC,  together  with  the  square  of  AC.  v.^.^.— > 

Upon    AB  describe  *  the  square  ADEB,  and  construct   the  a  46. 1. 
figure  as  in  the    preceding   propositions  :    and  because   AG  is 
equal  ^    to   GE,  add  to  each  of  them  CK  ;    the   whole  AK   is  b  43-  1- 
therefore    equal    to    the    whole    CE ; 
therefore     AK,    CE     are    double     of 
AK  :    but   AK,   CE   are   the   gnomon 
AKF  together  with  the    square    CK  ; 
therefore    the    gnomon     AKF,    toge- 
ther  with   the   square   CK,   is    double      Hi -p^ |K 

of  AK  :  but  twice  the  rectangle  AB, 
BC  is  double  of  AK,  for  BK  is  equal 
c  to  BC  :  therefore  the  gnomon  AKF, 
together  with  the  square  CK,   is  equal 

to  twice   the   rectangle    AB,    BC  :    to         D  F         E 

each  of  these  equals  add  HF,  which  is 

equal  to  the  square  of  AC  ;  therefore  the  gnomon  AKF,  toge- 
ther with  the  squares  CK,  HF,  is  equal  to  twice  the  rectangle 
AB,  BC,  and  the  square  of  AC  :  but  the  gnomon  AKF,  together 
with  the  squares  CK,  HF,  make  up  the  whole  figure  ADEB 
and  CK,  which  are  the  squares  of  AB  and  BC :  therefore  the 
squares  of  AB  and  BC  are  equal  to  twice  ihe  rectangle  AB,  BC, 
together  with  the  square  of  AC.  Wherefore,  if  a  straight  line, 
8cc.     Q.  E.  D, 


A              C 

B 

G 

/ 

X 

c  cor.  4.2. 


PROP.  VHI.     THEOR. 


IF  ia  straight  line  be  divided  into  any  two  parts, 
four  times  the  rectangle  contained  by  the  whole  line, 
and  one  of  the  parts,  together  with  the  square  of  the 
other  part,  is  equal  to  the  square  of  the  straight  line 
which  is  made  up  of  the  whole  and  that  part. 

Let  the  straight  line  AB  be  divided  into  any  two  parts  in  the 
point  C ;  four  times  the  rectangle  AB,  BC,  together  with  the 
square  of  AC,  is  equal  to  the  square  of  the  straight  line  m;ide 
up  of  AB  and  BC  together. 

Produce  AB  to  D,  so  that  BD  be  equal  to  CB,  and  upon 
AD  describe  the  square  AEFD  ;  and  construct  two  figures 
such  as  in  the  preceding.  Because  CB  is  equal  to  BD,  and 
that  CB   is   equal  *  to  GK,  and  BD  to  KN  ;   therefore  GK  is  a  34  1. 

H 


SB 


THE  ELEMENTS 


— — — — 1 

G 

I 

/ 

P 

/ 

R 

X 

Book  II.  equal  to  KN ;  for  the  same  reason,  PR  is  equal  to  RO  ;  and 
because  CB  is  equal  to  BD,  and  GK  to  KN,  the  rectangle 
CK  is  equal  •»  to  BN,  and  GR  to  RN  :  but  CK  is  equal  <=  to 
RN,^  because  they  are  the  complements  of  the  parallelogram. 
CO  ;  therefore  also  BN  is  equal  to  GR  ;  and  the  four  rect- 
angles BN,  CK,  GR,  RN  are  therefore  equal  to  one  another, 
and  so  are  quadruple  of  one  of  them  CK  :  again,  because  CB 
is  equal  to  BD,   and  that  BD  is 

dCor.4.2.  equal  ^  to   BK,  that  is,  to  CG  ;  C     B 

and  CB  equal  to  GK,  that  ^  is,  to      A , 1 1 yt  D 

GP ;    therefore   CG   is  equal   to 

GP  :  and  because  CG  is  equal  to     M I       ■     ..    •{ ^ — \  N 

GP,  and  PR  to  RO,  the  rectangle 

AG  is  equal  to   MP,   and  PL  to      X 7f — ! 1  O 

e  43. 1.  RF  :  but  MP  is  equal  «  to  PL, 
because  they  are  the  complements 
of  the  parallelogram  ML ;  where- 
fore   AG    is    equal    also    to    RF  :      ElZ^ :^ — J 1  F 

therefore      the      four     rectangles  "      ■'-' 

AG,  MP,  PL,  RF  are  equal  to 

one  another,  and  so  are  quadruple  of  one  of  them  AG.  And  it 
was  demonstrated,  that  the  four  CK,  BN,  GR,  and  RN  are  quad- 
ruple of  CK :  therefore  the  eight  rectangles  which  contain  the 
gnomon  AOH  are  quadruple  of  AK  :  and  because  AK  is  the 
rectangle  contained  by  AB,  BC,  for  BK  is  equal  to  BC,  four 
times  the  rectangle  AB,  BC  is  quadruple  of  AK ;  but  the  gno- 
mon AOH  was  demonstrated  to  be  quadruple  of  AK  ;  therefore 
four  times  the  rectangle  AB,  BC  is  equal  to  the  gnomon  AOH. 
To  each  of  these  add  XH,  which  is  equal  '^  to  the  square  of 
AC :  therefore  four  times  the  rectangle  AB,  BC,  together  with 
the  square  of  AC,  is  equal  to  the  gnomon  AOH  and  the  square 
XH  :  but  the  gnomon  AOH  and  XH  make  up  the  figure  AEFD, 
which  is  the  square  of  AD  :  therefore  four  times  the  rectangle 
AB,  BC,  together  with  the  square  of  AC,  is  equal  to  the  square 
of  AD,  that  is,  of  AB  and  BC  added  together  in  one  straight 
line.     Wherefore,  if  a  straight  line,  &c.     Q.  E.  D. 


OF  EUCLID. 


PROP.  IX.    THEOR. 


IF  a  straight  line  be  divided  into  two  equal,  and 
also  into  two  unequal  parts ;  the  squares  of  the  two 
unequal  parts  are  together  double  of  the  square  of 
half  the  line,  and  of  the  square  of  the  line  between 
the  points  of  section. 

Let  the  straight  line  AB  be  divided  at  the  point  C  into  two 
equal,  and  at  D  imo  two  unequal  parts  :  the  squares  of  AD, 
DB  are  together  double  of  the  squares  of  AC,  CD. 

From  the  point  C  draw  *  CE  at  right   angles  tb  AB,  and  a  11. 1 
make  it  equal  to  AC  or  CB,  and  join  EA,  EB ;  through  D  draw 
*>  DF  parallel  to  CE,  and  through  F  draw  FG  parallel  to  AB ;  b  31. 1. 
and  join  AF :   then,  because  AC  is  equal   to  CE,  the  angle 
EAC  is  equal   «=  to  the  angle  AEC  ;    and  because   the   angle  c  5. 1. 
ACE  is  a  right  angle,  the  two  others,   AEC,  EAC  together 
make  one  riglit  angle  ^ ;  and  they  are  equal  to  one  another ;  d  32. 1. 
each  of  them  therefoce   is   half  E 

of  a  right  angle.  For  the  same 
reason  each  of  the  angles  CEB, 
EBC  is  half  a  right  angle  ;  and 
therefore  the  whole  AEB  is  a 
right  angle  :  and  because  the  an- 
gle GEF  is  half  a  right  angle, 
and  EGF  a  right  angle,  for  it  is  A  CD  B 

equal  e  to  the  interior  and  oppo-  e  29. 1. 

site  angle  ECB,  the  remaining  angle  EFG  is  half  a  right  angle ; 
therefore  the  angle  GEF  is  equal  to  the  angle  EFG,  and  the 
side  EG  equal  ^  to  the  side  GF:  again,  because  the  angle  at  Bf  6. 1. 
js  half  a  right  angle,  and  FDB  a  right  angle,  for  it  is  equal 
^  to  the  interior  and  opposite  angle  ECB,  the  remaining  angle 
BFD  is  half  a  right  angle ;  therefore  the  angle  at  B  is  equal 
to  the  angle  BFD,  and  the  side  DF  to  ^  the  side  DB  :  and  be- 
cause AC  is  equal  to  CE,  the  square  of  AC  is  equal  to  the 
square  of  CE ;  therefore  the  squares  of  AC,  CE  are  double  of 
the  square  of  AC :  but  the  square  of  EA  is  equal  s  to  the  g  47. 1. 
squares  of  AC,  CE,  because  ACE  is  a  right  angle ;  therefore 
the  square  of  EA  is  double  of  the  square  of  AC:  again,  be- 
cause EG  is  equal  to  GF,  the  square  of  EG  is  equal  to  the 
2quare  of  GF ;  therefore  the  squares  of  FG,  GF  are  double  of 


60  THE  ELEMENTS 

Book  II.  the  square  of  GF  ;  but  the  square  of  EF  is  equal  to  the  squares 

'■"-V— ^  of  EG,  GF ;  therefore  the  square  of  EF  is  double  of  the  square 

h  34. 1.     GF ;  and  GF  is  equal  ^  to  CD  ;  therefore  the  square  of  EF  is 

double  of  the  square  of  CD  :  but  the  square  of  AE  is  likewise 

double  of  the  square  of  AC  ;  therefore  the  squares  of  AE,  EF 

are  double  of  the  squares  of  AC,  CD :  and  the  square  of  AF  is 

i  47. 1.     equal  '  to  the  squares  of  AE,  EF,  because  AEF  is  a  right  angle  ; 

therefore  the  square  of  AF  is   double  of  the   squares   of  AC, 

CD :   but  the  squares  of  AD,   DF  are   equal   to   the  square   of 

AF,  because  the  angle   ADF   ia  a  right  angle;   therefore   the 

squares  of  AD,   DF  are  double  of  the  squares  of  AC,  CD :   and 

DF  is  equal   to   DB;    therefore  the   squares    of   AD,   DB   are 

double  of  the  squares  of  AC,  CD.     If  therefore  a  straight  line, 

Sec.    Q.  E.  D. 


prop:  X.    THEOR. 


IF  a  straight  line  be  bisected,  and  produced  to  any 
point,  the  square  of  the  whole  line  thus  produced, 
and  the  square  of  the  part  of  it  produced,  are  together 
double  of  the  square  of  half  the  line  bisected,  and  of 
the  square  of  the  line  made  up  of  the  half  and  the 
part  produced. 

Let  the  straight  line  AB  be  bisected  in  C,  and  produced  to 
the  point  D  ;  the  squares  of  AD,  DB  are  double  of  the  squares 
of  AC,  CD. 

all.  1.  From  the  point  C  draw  ^  CE  at  right  angles  to  AB :  and 
make  it  equal  to  AC  or  CB,  and  join  AE,  EB  ;  through  E  draw 

b31. 1.  b  EF  parallel  to  AB,  i-.nd  through  D  draw  DF  parallel  to  CE : 
and  because  the  straight  line  EF  meets  the  parallels  EC,  FD,  the 

c  29. 1.  angles  CEF,  EFD  are  equal  <=  to  two  right  angles ;  and  therefore 
the  angles  BEF,  EFD  are  less  than  two  right  angles  :  but  straight 
lines  Avhich  with  another  straight  line  make  the  interior  angles 

d  12.  Ax.  upon  the  same  side  less  than  two  right  angles,  do  meet  'i  if  pro- 
duced far  enough :  therefore  EB,  FD  shall  meet,  if  produced, 
towards  B,  D  :  let  them  meet  in  G,  and  join  AG  :  then,  because 

e  5. 1.  AC  is  equal  to  CE,  the  angle  CEA  is  equal  e  to  the  angle 
EAC  ;  and  the  angle  ACE  is  a  right  angle  ;  therefore  each  of  the 

f  3S.  1.     angles  CEA,  EAC  is  half  a  right  angle  <":  for  the  same  reason. 


OF  EUCLID. 


61- 


h  34.  t. 


each  of  the  angles  CEB,  EBC  is  half  a  right  angle  ;  therefore  Book  II. 
AEB  is  a  right  angle  :  and  because  EBC  is  half  a  right  angle,  ^— -v— ^ 
DBG  is  also  *■  half  a  right  angle,  for  they  are  vertically  oppo-fl5.  1. 
site;  but  BDG  is  a  right  angle,  because  it  is  equal  ^  to  the  al-c29.  1. 
ternate  angle   DCE ;   therefore   the   remaining  angle    DGB    is 
half  a  right  angle,  and  is  therefore   equal  to  the  angle  DBG  ; 
wherefore  also  the  side  BD  is  equal  s  to  the  side  DG :  again,  g  6.  t. 
because   EGF  is   half  a 
right     angle,     and     that 
the  angle  at  F  is  a  right 
angle,    because   it   is    e- 
qual   ^    to    the    opposite 
angle  ECD,  the  remain- 
ing angle  FEG  is  half  a 
right    angle,    and    equal 
to     the      angle      EGF  ; 
wherefore    also   the    side 
GF  is  equal  s  to  the  side  FE.     And  because  EC  is  equal  to 
CA,  the  square  of  EC  is  equal  to  the  square  of  CA  ;    therefore 
the  squares  of  EC,   CA  are   double  of  the  square  of  CA  :  but 
the  square  of  EA  is  equal  »   to  the  squares  of  EC,  C A  ;  there-  i  47. 1. 
fore  the  square  of  EA  is  double  of  the  square  of  AC  :   again, 
because  GF  is  equal  to  FE,  the  square  of  GF  is  equal  to  the 
square  of  FE ;  and  therefore  the  squares  of  GF,  FE  are  dou- 
ble of  the  square  of  EF :  but  the   square  of  EG  is  equal  '  to 
the  squares  of  GF,  FE ;  therefore  the  square  of  EG  is  double 
of  the  square  of  EF :  and  EF  is  equal  to  CD  ;  wherefore  the 
square  of  EG  is  double  of  the  square  of  CD  :  but  it  was  demon- 
strated, that  the  square  of  EA  is  double  of  the  square  of  AC  ; 
therefore  the  squares  of  AE,  EG  are  double  of  the  squares  of 
AC,  CD  :    and  the  square  of  AG  is  equal  '  to  the  squares  of 
AE,  EG ;  therefore  the  square  of  AG  is  double  of  the  squares 
of  AC,  CD  :  but  the   squares  of  AD,   GD   are   equal  '■  to  the 
square  of  AG ;  therefore  the  squares  of  AD,  DG  are  double  of 
the  squares  of  AC,  CD  :   but  DG  is  equal  to  DB  ;  therefore  the 
squares  of  AD,    DB   are   double  of  the   squares  of  AC,  CD. 
Wherefore,  if  a  straight  Ihie,  &c.     Q.  E.  D. 


THE  ELEMENTS 


PROP.  XI.  PROB, 


a  46.  1. 
b  10. 1. 
c  3.  1. 


d6.  2. 


e  47.  1. 


TO  divide  a  given  straight  line  into  two  parts,  so 
that  the  rectangle  contained  by  the  whole  and  one  of 
the  parts  shall  be  equal  to  the  square  of  the  other 
part. 

Let  AB  be  the  given  straight  line ;  it  is  required  to  divide  it 
into  tv/o  parts,  so  that  the  rectangle  contained  by  the  whole  and 
one  of  tlie  parts  shall  be  equal  to  the  square  of  the  other  part. 

Upon  AB  describe  ^  the  square  ABDC  ;  bisect  b  AC  in  E,  and 
join  BE;  produce  CA  to  F,  and  make  «=  EF  equal  to  EB  ;  and 
upon  AF  describe  »  the  square  FGHA  ;  AB  is  divided  in  H,  so 
that  the  rectangle  AB,  BH  is  equal  to  the  square  of  AH. 

Produce  GH  to  K :  because  the  straight  line  AC  is  bisected 
in  E,  and  produced  to  the  point  F,  the  rectangle  CF,  FA,  to- 
gether Vv'ith  the  square  of  AE,  is  equal  ^  to  the  square  of  EF  : 
but  EF  is  equal  to  EB ;  therefore  the  rectangle  CF,  FA,  toge- 
ther with  the  square  of  AE,  is  equal  to  the  square  of  EB  :  and  the 
squares  of  BA,  AE  are  equal  e  to  the         F  G 

square  of  EB,  because  the  angle  EAB 
is  a  right  angle  ;  therefore  the  rect- 
angle CF,  FA,  together  with  the  square 
of  AE,  is  equal  to  the  squares  of  BA, 
AE :  take  away  the  square  of  AE, 
"which  is  common  to  both,  therefore 
the  remaining  rectangle  CF,  FA  is 
equal  to  the  square  of  AB  :  and  the  fi- 
gure FK  is  the  rectangle  contained  by 
CF,  FA,  for  AF  is  equal  to  FG ;  and 
AD  is  the  square  of  AB ;  therefore 
FK  is  equal  to  AD :  take  away  the 
common  part  AK,  and  the  remainder 
FH    is    equal    to   the    remainder  HD :         C  K         D 

and  HD  is  the  rectangle  contained  by  AB,  BH,  for  AB  is  equal 
to  BD ;  and  FH  is  the  square  of  AH :  therefore  the  rectangle 
AB,  BH  is  equal  to  the  square  of  AH :  wherefore  the  straight 
lirie  AB  is  divided  in  H  so,  that  the  rectangle  AB,  BH  is  equal  to 
the  square  of  AH.     Which  was  to  be  done. 


OF  EUCLID. 


PROP.  XII.    THEOR. 


IN  obtuse  angled  triangles,  if  a  perpendicular  be  * 
drawn  from  any  of  the  acute  angles  to  the  opposite 
side  produced,  the  square  of  the  side  subtending  the 
obtuse  angle  is  greater  than  the  squares  of  the  sides 
containing  the  obtuse  angle,  by  twice  the  rectangle 
contained  by  the  side  upon  which,  when  produced, 
the  perpendicular  falls,  and  the  straight  line  inter- 
cepted without  the  triangle  between  the  perpendicu- 
lar and  the  obtuse  angle. 

Let  ABC  be  an  obtuse   angled  triangle,  having  the   obtuse 
angle  ACB,  and  from  the  point  A  let  AD  be  drawn  ^  perpendi-  a  12.  1. 
cular  to  BC  produced:   the  square  of  AB  is  greater  than  the 
squares  of  AC,  CB  by  twice  the  rectangle  BC,  CD. 

Because  the  straight  line  BD  is  divided  into  two  parts  in  the 
point  C,  the  square  of  BD  is  equal 

^  to  the  squares  of  BC,  CD,  and  ^  A  b  4.  2. 

twice  the  rectangle  BC,  CD :  to 
each  of  these  equals  add  the  square 
of  DA ;  and  the  squares  of  BD,  DA 
ai'e  equal  to  the  squares  of  BC,  CD, 
DA,  and  twice  the  rectangle  BC, 
CD  :  but  the  square  of  B  A  is  equal 

<^  to  the  squares  of  BD,  DA,  be-        y  /  |       c  47. 1. 

cause  the   angle  at  D  is  a  right  -d'' 
angle ;  and  the   square  of  C  A  is 

equal  «  to  the  squares  of  CD,  DA :  therefore  the  square  of  B A 
is  equal  to  the  squares  of  BC,  CA,  and  twice  the  rectangle  BC, 
CD  ;  that  is,  the  square  of  BA  is  greater  than  the  squares  of 
BC,  CA,  by  twice  the  rectangle  BC,  CD.  Therefore,  in  obtuse 
angled  triangles,  &c.     Q.  E.  D. 


64 
Book  II. 


THE  ELEMENTS 


PROP.  XIII.    THEOR. 


and  because 


See  N.  IN  every  triangle,  the  square  of  the  side  subtend-- 
ing  any  of  the  acute  angles  is  less  than  the  squares  of 
the  sides  containing  that  angle,  by  twice  the  rectangle 
contained  by  either  of  these  sides,  and  the  straight  line 
intercepted  between  the  perpendicular  let  fall  upon  it 
from  the  opposite  angle,  and  the  acute  angle. 

Let  ABC  be  any  triangle,  and  the  angle  at  B  one  of  its  acute 
angles,  and  upon  BC,  one  of  the  sides  containing  it,  let  fall  the 

a  12. 1.  perpendicular  *  AD  from  the  opposite  angle :  the  square  of 
AC,  opposite  to  the  angle  B,  is  less  than  the  squares  of  CB,  BA, 
by  twice  the  rectangle  CB,  BD. 

First,  Let   AD  fall   within   the   triangle  ABC; 
the  straight  line  CB  is  divided 
into  two  parts  in  the   point  D, 
the  squares  of  CB,  BD  are  equal 

b  7.  12  ^  to  twice  the  rectangle  contain- 
ed by  CB,  BD,  and  the  square 
of  DC:  to  each  of  these  equals 
add  the  square  of  AD  ;  therefore 
the  squares  of  CB,  BD,  DA  are 
equal  to  twice  the  rectangle  CB, 
BD,  and  the  squares  of  AD,  DC  : 

<;  47-  1     but  the  square  of  AB  is  equal  =  to         B  D 

the  squares  of  BD,  DA,  because  the  angle  BDA  is  a  right  angle  ; 
and  the  square  of  AC  is  equal  to  the  squares  of  AD,  DC  : 
therefore  the  squares  of  CB,  BA  are  equal  to  the  square  of  AC, 
and  twice  the  rectangle  CB,  BD,  that  is,  the  square  of  AC  alone 
is  less  than  the  squares  of  CB,  BA  by  twice  the  rectangle  CB, 
BD. 

Secondly,  Let  AD   fall  with-  A 

out  the  triangle  ABC :  then,  be- 
cause the  angle  at  D  is  a  right 
angle,  the  angle  ACB  is  greater 

(1 16.  1.    'I  ihan  a  right  angle  ;   and  there- 

e  12.  2.     fore  the  square  of  AB  is  equal'^  to 

the  squares  of  AC,  CB,  and  twice 

the  rectangle  BC,  CD  :  to  these  e- 

qualsaddthe  squareofBC,andthe 

BCD 


OF  EUCLID. 


65 


squares  of  AB,  BC  are  equal  to  the  square  of  AC,  and  twice  Book  ir, 
the  square  of  BC,  and  twice  the   rectangle  BC,  CD :    but  be-  ^■— v— ^ 
cause  BD  is  divided  into  two  parts  in  C,  the  rectangle  DB,  BC 
is  equal  f  to  the  rectangle  BC,  CD  and  the  square  of  BC  :    and  f  3.  2. 
the  doubles  of  these  are  equal :    therefore  the  squares  of  AB, 
BC  are  equal  to  the  square  of  AC,   and  twice  the  rectangle 
DB,  BC :    therefore  the  square  of  AC  alone   is  less  than  the 
squares  of  AB,  BC  by  twice  the  rectangle  DB,  BC.  A 

Lastly,  Let  the  side  AC  be  perpendicular  to 
BC ;  then  is  BC  the  straight  line  between  the 
perpendicular  and  the  acute  angle  at  B ;  and  it 
is  manifest  that  the  squares  of  AB,  BC  are  equal 
K  to  the  square  of  AC  and  twice  the  square  of  BC. 
Therefore,  in  every  triangle,  &:c.    Q.  E.  D. 


.     PROP.  XIV.    PROB. 


TO  describe  a  square  that  shall  be  equal  to  a  given  See  n. 
rectilineal  figure. 

Let  A  be  the  given  rectilineal  figure  ;  it  is  required  to  describe 
a  square  that  shall  be  equal  to  A. 

Describe  *  the  rectangular  parallelogram  BCDE  equal  tt)  the  a  45. 1. 
rectilineal  figure  A.     If,  then,  the  sides  of  it  BE,  ED  are  equal 

to  one  another,  it  a  ^ ^.^^^   jr 

is  a  square,  and 
what  was  requir- 
ed is  now  done  : 
but  if  they  are  not 
equal,  produce  one   \  /   •' B 

of  them  BE  to  F, 
and  make  EF  e- 
qual  to  ED,  and 
bisect  BF  in  G ; 
and  from  the  centre  G,  at  the  distance  GB,  or  GF,  describe  the 
semicircle  BHF,  and  produce  DE  to  H,  and  join  GH  ;  therefore, 
because  the  straight  line  BF  is  divided  into  two  equal  parts  in 
the  point  G,  and  into  two  unequal  at  E,  the  rectangle  BE, 
EF,  together  with  the  square  of  EG,  is  equal  ^  to  the  square  ofb  5,  2. 
GF :  but  GF  is  equal  to  GH ;  therefore  the  rectangle  BE,  EF, 

I 


66  THE  ELEMENTS,  Sec. 

Book  II.  together  with  the  square  of  EG,  is  equal  to  the  square  of  GH ; 
*--"v-— ^  but  the  squares  of  HE,  EG  are  equal  «  to  the  square  of  GH : 
c  47. 1.  therefore  the  rectangle  BE,  EF,  together  with  the  square  of 
EG,  is  equal  to  the  squares  of  HE,  EG:  take  away  the  square 
of  EG,  which  is  common  to  both ;  and  the  i^emaining  rect- 
angle BE,  EF  is  equal  to  the  square  of  EH :  but  the  rectangle 
contained  by  BE,  EF  is  the  parallelogram  BD,  because  EF  is 
equal  to  ED  ;  therefore  BD  is  equal  to  the  square  of  EH  ;  but 
BD  is  equal  to  the  rectilineal  figure  A ;  therefore  the  rectilineal 
figure  A  is  equal  to  the  square  of  EH :  wherefore  a  square  has 
been  made  equal  to  the  given  rectilineal  figure  A,  viz.  the  square 
described  upon  EH.     Which  was  to  be  done. 


THE 


ELEMENTS  OF  EUCLID- 


BOOK  III. 


DEFINITIONS. 


I. 

JljQUAL  circles  are  those  of  which  the  diameters  are  equal,  or  Book  III. 

from  the  centres  of  which  the  straight  lines  to  the  circumfer-  s— ^-—^ 

ences  are  equal. 

'  This  is  not  a  definition  but  a  theorem,  the  truth  of  which  is 
'  evident ;  for,  if  the  circles  be  applied  to  one  another,  so  that 
'  their  centres  coincide,  the  circles  must  likewise  coincide,  since 
*  the  straight  lines  from  the  centres  are  equal.' 

II. 
A  straight  line  is  said  to  touch 

a  circle,  when  it  meets  the 

circle,    and   being   produced 

does  not  cut  it. 

III. 
Circles  are  said  to  touch  one 

another,  which  meet,  but  do 

not  cut  one  another. 
IV. 
Straight  lines  are  said  to  be  equally  distant 

from  the  centre  of  a  circle,  when  the 

perpendiculars  drawn  to  them  from  the 

centre  are  equal. 
V. 
And  the  straight  line  on  which  the  greater 

perpendicular  falls,  is  said  to  be  farther 

from  the  centre. 


68 


THE  ELEMENTS 


Book  III-  VI. 

^■■^■v^*^  A  segment  of  a  circle  is  the  figure  con- 
tained by  a  straight  line  and  the  cir- 
cumference it  cuts  off. 
VII. 
"  The  angle  of  a  segment  is  that  which  is  contained  by  the 
"  straight  line  and  the  circumference." 
VIII. 
An  angle  in  a  segment  is  the  angle  con- 
tained by   two   straight   lines   drawn 
from  any  point  in  the  circumference 
of  the  segment,  to  the  extremities  of 
the  straight  line  which  is  the  base  of 
the  segment. 

IX. 
And  an  angle  is  srid  to  insist  or  stand 
upon   the   circumference    intercepted 
between  the  straight  lines  that  contain 
the  angle. 

X. 
The  sector  of  a  circle  is  the  figure  contain- 
ed by  two  straight  lines  drawn  from  the 
centre,  and   the  circumference  between 
them. 

XI. 

Similar  segments  of  a  circle 
are  those  in  which  the  an- 
gles are  equal,  or  which 
contain  equal  angles. 


PROP.  I.    PROB. 


See  N.         To  find  the  centre  of  a  given  circle. 


a  10.  1. 
bll.  1. 


Let  ABC  be  the  given  circle  ;   it  is  required  to  find  its  centre. 

Draw  within  it  any  straight  line  AB,  and  bisect  »  it  in  D  ; 
from  the  point  D  draw  ^  UC  at  rjgh.t  angles  to  AB,  and  pro- 
duce it  to  E,  and  bisect  CE  in  F :  the  point  F  is  the  centre  of  the 


circle  ABC. 


OF  EUCLID, 


69 


For,  if  it  be  not,  let,  if  possible,  G  be  the  centre,  and  join    B.  Ill, 
GA,  GD,  GB  :    then,  because  DA  is  equal  to  DB,  and  DG  *— v— 
common  to  the  two  triangles  ADG, 
BDG,  the  two  sides  AD,  DG  are  e-  G 

qual  to  the  two  BD,  DG,  each  to 
each  ;  and  the  base  GA  is  equal  to 
the  base  GB,  because  they  are  drawn 
from  the  centre  G  * :  therefore  the 
angle  ADG  is  equal  <=  to  the  angle 
GDB  :  but  when  a  straight  line  stand- 
ing upon  another  straight  line  makes 
the  adjacent  angles  equal  to  one  ano- 
ther, each  of  the  angles  is  a  right  an- 
gle <i :  therefore  the  angle  GDB  is  a 
right  angle :  but  FDB  is  likewise  a  " 

right  angle ;  wherefore  the  angle  FDB  is  equal  to  the  angle 
GDB,  the  greater  to  the  less,  which  is  impossible :  therefore  G 
is  not  the  centre  of  the  circle  ABC :  in  the  same  manner  it  can 
be  shown,  that  no  other  point  but  F  is  the  centre ;  that  is,  F  is 
the  centre  of  the  circle  ABC.     Which  was  to  be  found. 

Cor.  From  this  it  is  manifest,  that  if  in  a  circle  a  straight 
line  bisect  another  at  right  angles,  the  centre  of  the  circle  is  in 
the  line  which  bisects  the  other. 


c8. 1. 


d  10.  dcf, 
1. 


PROP.   II.     THEOR. 

IF  any  two  points  be  taken  in  the  circumference  of 
a  circle,  the  straight  line  which  joins  them  shall  fall 
within  the  circle. 


Let  ABC  be  a  circle,  and  A,  B  any  two  points  in  the  circum- 
ference ;  the  straight  line  drawn  from  C 
A  to  B  shall  fall  within  the  circle. 

For,  if  it  do  not,  let  it  fall,  if  possi- 
ble, without,  as  AEB  ;  find  ^  D  the  cen- 
tre of  the  circle  ABC,  and  join  AD, 
DB,  and  produce  DF,  any  straight  line 
meeting  the  circumference  AB,  to  E  : 
then  because  DA  is  equal  to  DB,  the 
angle  DAB  is  equal  ^  to  the  angle  DB  A ; 
and  because  AE,  a  side  of  the  triangle 

*  N.  B.  Whenever  the  expression  "  straight  lines  from  the  centre,"  or 
"  drawn  from  the  centre,"  occurs,  it  is  to  be  understood  that  they  are  drawn 
to  the  circumference. 


al.S. 


b5.  1. 


70 


THE  ELEMENTS 


Booklll.  DAE,  is  produced  to  B,  the  angle  DEB  is  greater  ^  than  the 
■  angle  DAE:  but  DAE  is  equal  to  the  angle  DBE ;  therefore 
the  angle  DEB  is  greater  than  the  angle  DBE :  but  to  the  great- 
er angle  the  greater  side  is  opposite  ^ ;  DB  is  therefore  greater 
than  DE:  but  DB  is  equal  to  DF  ;  wherefore  DF  is  greater 
than  DE,  the  less  than  the  greater,  which  is  impossible :  there- 
fore the  straight  line  drawn  from  A  to  B  does  not  fall  without 
the  circle.  In  the  same  manner  it  may  be  demonstrated  that  it 
does  not  f«Il  upon  the  circumference  ;  it  falls  therefore  within  it. 
Wherefore,  if  any  two  points,  Sec.     Q.  E.  D. 


PROP.  III.    THEOR. 


IF  a  straight  line  drawn  through  the  centre  of  a 
circle  bisect  a  straight  line  in  it  which  does  not  pass 
through  the  centre,  it  shall  cut  it  at  right  angles  ;  and, 
if  it  cuts  it  at  right  angles,  it  shall  bisect  it. 


Let  ABC  be  a  circle  ;  and  let  CD,  a  straight  line  drawn 
through  the  centre,  bisect  any  straight  line  AB,  which  does  not 
pass  through  the  centre,  in  the  point  F  :  it  cuts  it  also  at  right 
angles. 

3^  1.  3.  Take  ^  E  the  centre  of  the  circle,  and  join  EA,  EB.     Then, 

because  AF  is  equal  to  FB,  and  FE  common  to  the  two  tri- 
angles AFE,  BFE,  there  are  two  sides  in  the  one  equal  to  two 
sides  in  the  other,    and  the  base  EA  is  C 

equal  to  the  base  EB  ;  therefore  the  angle 

b  8. 1.  AFE  is  equal  ^  to  the  angle  BFE :  but 
when  a  straight  line  standing  upon  ano- 
ther makes  the  adjacent  angles  equal  to 

c  10.  def.  one  anotl>er,  each  of  them  is  a  right  «  an- 
^  gle :  therefore  each  of  the  angles  AFE, 

BFE  is  a  right  angle  ;  wherefore  the 
straight  line  CD,  drawn  through  the  cen- 
tre bisecting  another  AB  that  does  not 
pass  through  the  centre,  cuts  the  same  at 
right  angle:;. 

But  let  CD  cut  AB  at  right  angles  j  CD  also  bisects  it,  that 
is,  AF  is  equal  to  FB. 

The  same  construction  being  made,  because  EA,  EB  from 

<\5A.  the  centre  a'e  equal  to  one  another,  the  angle  EAF  is  equal  «l 
to  the  angle  EBF ;  and  the  right  angle  AFE  is  equal  to  the 
right  angle  BFE  ;  therefore,  in  the  two  triangles  EAF,  EBF, 


OF  EUCLID.  7i 

there  are  two  angles  in  one  equal  to  two  angles  in  the  other,  Book  III, 
and  the  side  EF,  which  is  opposite  to  one  of  the  equal  angles  v  —^..^ 
in  each,  is  common  to  both  ;  therefore  the  other  sides  are  equal  e  ;  e  26. 1. 
AF  therefore  is  equal  to  FB.     Wherefore,  if  a  straight  line,  &c. 
Q.  E.  D. 


PROP.  IV.    THEOR. 

IF  in  a  circle  two  straight  lines  cut  one  another 
which  do  not  both  pass  through  the  centre,  they  do 
not  bisect  each  other. 

Let  ABCD  be  a  circle,  and  AC,  BD  two  straight  lines  in  it 
•which  cut  one  another  in  the  point  E,  and  do  not  both  pass  through 
the  centre  ;  AC,  BD  do  not  bisect  one  another. 

For,  if  it  is  possible,  let  AE  be  equal  to  EC,  and  BE  to  ED  : 
if  one  of  the  lines  pass  through  the  centre,  it  is  plain  that  it 
cannot  be  bisected  by  the  other  which 
does  not  pass  through  the  centre :  but, 
if  neither  of  them  pass   through  the 

centre,  take  *  F  the  centre  of  the  circle,      /  p  \D  dii.3. 

and    join    EF :    and    because   FE,    a 
straight  line  through  the  centre,  bisects  -^ 
another    AC    which    does    not    pass 
through   the  centre,  it  shall  cut  it  at 
right  •»  angles  ;    wherefore  FEA  is  a  B  "^ — — -^^    C       b  3.  3, 

right  angle :  again,  because  the  straight  line  FE  bisects  the 
straight  line  BD  which  does  not  pass  through  the  centre,  it  shall 
cut  it  at  right  bangles:  wherefore  FEB  is  a  right  angle  :  and  FEA 
was  shown  to  be  a  right  angle ;  therefore  FEA  is"  equal  to  the 
angle  FEB,  the  less  to  the  greater,  which  is  impossible :  there- 
fore AC,  BD  do  not  bisect  one  another.  Wherefore,  if  in  a  cir- 
cle, &c.    Q.  E.  D. 


PROP.  V.     THEOR. 

IF  two  circles  cut  one  another,  they  shall  not  have 
the  same  centre. 

Let  the  two  circles  ABC,  CDG  cut  one  another  in  the  points 
B,  C ;  they  have  not  the  same  centre. 


72 


THE  ELEMENTS 


Booklll.      For,  if  it  be  possible,  let  E  be  their  centre:    join  EC,  and 

^"■^v*^  draw  any  straight  line  EFG  meet- 
ing them  in  F  and  G  ;  and  because  C 
E  is  the  centre  of  the  circle  ABC, 
CE  is  equal  to  EF :  again,  be- 
cause E  is  the  centre  of  the  circle 
CDG,  CE  is  equal  to  EG:  but  ^ 
CE  was  shown  to  be  equal  to  EF ; 
therefore  EF  is  equal  to  EG,  the 
less  to  the  greater,  which  is  impos- 
sible :  therefore  E  is  not  the  centre 
of  the  circles  ABC,  CDG.  Where- 
fore, if  two  circles,  &c.     Q.  E.  D. 


PROP.  VI.     THEOR. 


IF  two  circles  touch  one  another  internally,  they 
shall  not  have  the  same  centre. 

Let  the  two  circles  ABC,  CDE  touch  one  another  internally  in 
in  the  point  C ;  they  have  not  the  same  centre. 

For,  if  they  can,  let  it  be  F ;   join  FC,  and  draw  any  straight 
line  FEB  meeting  them  in  E  and  B;  C 

and  because  F  is  the  centre  of  the 
circle  ABC,  CF  is  equal  to  FB ; 
also,  because  F  is  the  centre  of  the 
circle  CDE,  CF  is  equal  to  FE :  and 
CF  was  shown  equal  to  FB ;  there- 
fore FE  is  equal  to  FB,  the  less  to  A 
the  greater,  which  is  impossible: 
wherefore  F  is  not  the  centre  of 
the  circles  ABC,  CDE.  Therefore, 
if  two  circles,  Stc.     Q.  E.  D. 


OF   EUCLID. 


PROP.  VII.    THEOR, 


IF  any  poir>t  be  taken  in  the  diameter  of  a  circle, 
which  is  not  the  centre,  of  all  the  straight  lines  which 
can  be  drawn  from  it  to  the  circumference,  the  greatest 
is  that  in  which  the  centre  is,  and  the  other  part  of 
that  diameter  is  the  least ;  and,  of  any  others,  that 
which  is  nearer  to  the  line  which  passes  through  the 
centre  is  always  greater  than  one  more  remote  :  and 
from  the  same  point  there  can  be  drawn  only  two 
straight  lines  that  are  equal  to  one  another,  one  upon 
each  side  of  the  shortest  line. 

Let  ABCD  be  a  circle,  and  AD  its  diameter,  in  which  let 
any  point  F  be  taken  which  is  not  the  centre;  let  the  centre 
be  E  ;  of  all  the  straight  lines  FB,  FC,  FG,  Sec.  that  can  be  drawn 
from  F  to  the  circumference,  FA  is  the  greatest,  and  FD,  the 
other  part  of  the  diameter  AD,  is  the  least:  and  of  the  others, 
FB  is  greater  than  FC,  and  FC  than  FG. 

Join  BE,  CE,  GE ;  and  because  two  sides  of  a  triangle  are 
greater  a  than  the  third,  BE,  EF  are  greater  than  BF;  but  AE*20. !.. 
is  ^equal  to  EB ;  therefore  AE,  EF, 
that  is,  AF,  is  greater  than  BF : 
again,  because  BE  is  equal  to  CE, 
and  FE  common  to  the  triangles 
BEF,  CEF,  the  two  sides  BE,  EF 
are  equal  to  the  two  CE,  EF;  but 
the  angle  BEF  is  greater  than  the 
angle  CEF  ;  therefore  the  base  BF  is  ^^ 

greater »» than  the  base  FC :   for  the        \     //ivN^  /         b24. 1, 
same  reason,  CF  is  greater  than  GF : 
again,  because  GF,  FE  are  greater 
*  than   EG,    and    EG    is   equal    to 

ED;  GF,  FE  are  greater  than  ED:  take  away  the  comTnon 
part  FE,  and  the  remainder  GF  is  greater  than  the  remainder 
FD :  therefore  FA  is  the  greatest,  and  FD  the  least  of  all  the 
straight  lines  from  F  to  the  circumference ;  and  BF  is  greater 
than  CF,  and  CF  than  GF. 

Also  there  can  be  drawn  only  two  equal  straight  lines  from 
ihe  point  F  to  the  circumference,  one  upon  eaeh  side  of  the 

K 


r*  THE  ELEMENTS 

Book  in,  shortest  line  FD :  at  the  point  E,  in  the  straight  line  EF,  make 
^•^-v-*^  c  the  angle  FEH  equal  to  the  angle  GEF,  and  join  FH :    then 
c  23.  1.    because  GE  is  equal  to  EH,  and  EF  common  to  the  two  tri- 
angles GEF,  HEF ;   the  two  sides  GE,  EF  are  equal  to  the  two 
HE,  EF ;  and  the  angle  GEF  is  equal  to  the  angle  HEF  ;  there- 
(j  4. 1,      fore  the  base  EG  is  equal  ^  to  the  base  FH  :   but,  besides  FH,  no 
other  straight  line  can  be  drawn  from  F  to  the  circumference 
equal  to   FG :    for,  if  there   can,  let  it  be   FK ;    and   because 
FK  is  equal  to  FG,  and  FG  to  FH,  FK  is  equal  to  FH ;    that  is, 
a  line  nearer  to  that  which  passes  through  the  centre,  is  equal  to 
one  which  is  more  remote;  which  is  impossible.     Therefore,  if 
any  point  be  taken,  Sec.     Q.  E.  D. 


PROP.  VHI.    THEOR. 


IF  any  point  be  taken  without  a  circle,  and  straight 
lines  be  drawn  from  it  to  the  circumference,  where- 
of one  passes  through  the  centre,  of  those  which  fall 
upon  the  concave  circumference,  the  greatest  is  that 
which  passes  through  the  centre,  and,  of  the  rest,  that 
^vhich  is  nearer  to  that  through  the  centre  is  always 
greater  than  the  more  remote  :  but  of  those  which  fall 
upon  the  convex  circumference,  the  least  is  that  be- 
tween the  point  without  the  circle,  and  the  diameter; 
and,  of  the  rest,  that  which  is  nearer  to  the  least  is 
always  less  than  the  more  remote  :  and  only  two  equal 
straight  lines  can  be  drawn  from  the  point  unto  the 
circumference,  one  upon  each  side  of  the  least. 


Let  ABC  be  a  circle,  and  D  any  point  without  it,  from  which 
let  the  straight  lines  DA,  DE,  DF,  DC  be  drawn  to  the  cir- 
cumference, whereof  DA  passes  through  the  centre.  Of  those 
which  fall  upon  the  concave  part  of  the  circumference  AEFC, 
the  greatest  is  AD  which  passes  through  the  centre  ;  and  the 
nearer  to  it  is  always  greater  than  the  more  remote,  viz.  DE 
than  DF,  and  DF  than  DC ;  but  of  those  which  fall  upon  the 
convex  circumference  HLKG,  the  least   is   DG   between  the 


OF  EUCLID. 


75 


d  4.  Ax. 


point  D  and  the  diameter  AG;  and  the  nearer  to  it  is  always  Book  III 
less  than  the  more  remote,  viz.  DK  than  DL,  and  DL  than  *— y— ^ 
DH. 

Take  »  M  the  centre  of  the  circle  ABC,  and  join  ME,  MF, » 1.  3. 
MC,  MK,  ML,  MH:   and  because  AM  is  equal  to  ME,  add 
MD  to  each,  therefore  AD  is  equal  to  EM,  MD ;  but  EM,  MD 
are  greater^  than  ED;  therefore  also  AD  is  greater  than  ED  :  b  20. 1, 
again,  because  ME  is  equal  to  MF,  and  MD  common  to  the 
triangles   EMD,   KMD  ;   EM,   MD 
are   equal   to    FM,    MD  ;    but    the 
angle    EMD    is    greater    than    the 
angle     FMD  ;    therefore    the    base 

ED  is  greater '^  than  the  base  FD  :  ///  \\  c24. 1, 
in  like  manner  it  may  be  shown 
that  FD  is  greater  than  CD  : 
therefore  DA  is  the  greCest :  and 
DE  greater  than  DF,  and  DF  than 
DC:  and  because  MK,  KD  are 
greater  •>  than  MD,  and  MK  is 
equal  to  MG,  the  remainder  KD 
is  greater  '^  than  the  remainder 
GD,  that  is,  GD  is  less  than  KD  : 
and  because  MK,  DK  are  drawn 
to  the  point  K  within  the  triangle 
MLD  from  M,  D,  the  extremi- 
ties of  its  side  MD ;  MK,  KD  are                      

less  e  than  ML,  LD,  whereof  MK  E        A  e  21- 1. 

is  equal  to  ML ;  therefore  the  remainder  DK  is  less  than  the  re- 
mainder DL :  in  like  manner  it  may  be  shown,  that  DL  is  less 
than  DH  :  therefore  DG  is  the  least,  and  DK  less  than  DL,  and 
DL  than  DH  :  also  there  can  be  drawn  only  two  equal  straight 
lines  from  the  point  D  to  the  circumference,  one  upon  each 
side  of  the  least :  at  the  point  M,  in  the  straight  line  MD,  make 
the  angle  DMB  equal  to  the  angle  DMK,  and  join  DB  :  and 
because  MK  is  equal  to  MB,  and  MD  common  to  the  triangles 
KMD,  BMD,  the  two  sides  KM,  MD  are  equal  to  the  two  BM, 
MD  ;  and  the  angle  KMD  is  equal  to  the  angle  BMD  ;  there- 
fore the  base  DK  is  equal  f  to  the  base  DB  :  but,  besides  DB,  f  4. 1. 
there  can  be  no  straight  line  drawn  from  D  to  the  circumference 
equal  to  DK  :  for,  if  there  can,  let  it  be  DN  ;  and  because  DK  is 
equal  to  DN,  and  also  to  DB  ;  therefore  DB  is  equal  to  DN,  that 
is,  the  nearer  to  the  least  equal  to  the  more  remote,  which  is 
impossible.     If,  therefore,  any  point,  &c.     Q.  E,  D. 


THE  ELEMENTS 


PROP.  IX.    PROB. 


a  7.  3. 


IF  a  point  be  taken  within  a  circle,  from  which 
there  fall  more  than  two  equal  straight  lines  to  the 
circumference,  that  point  is  the  centre  of  the  circle. 

Let  the  point  D  be  taken  within  the  circle  ABC,  from 
Avhich  to  the  circumference  there  fall  more  than  two  equal 
straight  lines,  viz.  DA,  DB,  DC ;  the  point  D  is  the  centre  of 
the  circle. 

For,  if  not,  let  E  be  the  centre, 
join  DE  and  produce  it  to  the  cir- 
cumference in  F,  G ;  then  FG  is 
a  diameter  of  the  circle  ABC  :  and 
because  in  FG,  the  diameter  of  the 
circle  ABC,  there  is  taken  the 
point  D  which  is  not  the  centre,  DG 
shall  be  the  greatest  line  from  it  to 
the  circumference,  and  DC  greater 
a  than  DB,  and  DB  than  DA  ;  but 
they  are  likewise  equal,  which  is 
impossible  :  therefore  E  is  not  the 

centre  of  the  circle  ABC  :  in  like  manner,  it  may  be  demon- 
strated, that  no  other  point  but  D  is  the  centre ;  D  therefore  is 
the  centre.     Wherefore,  if  a  point  be  taken,  &c.     Q.  E.  D. 


a  9.  3. 


PROP.  X.    THEOR. 

ONE  circumference  of  a  circle  cannot  cut  another 
in  more  than  two  points. 


If  it  be  possible,  let  the  circumfe- 
rence FAB  cut  the  circumference 
DEF  in  more  than  two  points,  viz. 
in  B,  G,  F ;  take  the  centre  K  of  the 
circle  ABC,  and  join  KB,  KG,  KF: 
and  because  within  the  circle  DEF 
there  is  taken  the  point  K,  from 
which  to  the  circumference  DEF 
fall  more  than  two  equal  straight 
lines  KB,  KG,  KF,  the  point  K  is» 


OF  EUCLID.  77 

the  centre  of  the  circle  DEF :  but  K  is  also  the  centre  of  the  Book  III. 
circle  ABC ;  therefore  the  same  point  is  the  centre  of  two  cir    ^  ."r-mj 
cles  that  cut  one  another,  which  is  impossible''.    Therefore  one  b 5.  3. 
circumference  of  a  circle  cannot  cut  another  in  more  than  two 
points.     Q.  E.  D. 


PROP.  XL    THEOR. 

IF  two  circles  touch  each  other  internally,  the 
straight  line  which  joins  their  centres  being  produced 
shall  pass  through  the  point  of  contact. 

Let  the  two  circles  ABC,  ADE  touch  each  other  internally 
in  the  point  A,  and  let  F  be  the  centre  of  the  circle  ABC,  and 
G  the  centre  of  the  circle  ADE ;  the  ^ 

straight  line  which  joins  the  centres 
F,  G,  being  produced,  passes  through 
the  point  A. 

For,  if  not,  let  it  fall  otherwise,  if  pos^ 
sible,  as  FGDH,    and  join  AF,  AG : 

and  because  AG,  GF  are  greater^  than  |y  ^  7?\^  /  i  ^20.1- 
FA,  that  is,  than  FH,  for  FA  is  equal  to 
FH,  both  being  from  the  same  centre ; 
take  away  the  common  part  FG ;  there- 
fore the  remainder  AG  is  greater  than 
the  remainder  GH  :  but  AG  is  equal 

to  GD ;  therefore  GD  is  greater  than  GH,  the  less  than  the 
greater,  which  is  impossible.  Therefore  the  straight  line  which 
joins  the  points  F,  G  cannot  fall  otherwise  than  upon  the  point 
A,  that  is,  it  must  pass  through  it.  Therefore,  if  two  circles,  5cg. 
Q.  E.  D. 


PROP.  XIL    THEOR. 

IF  two  circles  touch  each  other  externally,  the 
straight  line  which  joins  their  centres  shall  pass 
through  the  point  of  contact. 

Let  the  two  circles  ABC,  ADE  touch  each  other  externally 
in  the  point  A  ;  and  let  F  be  the  centre  of  the  ciix;le  ABC,  and 
G  the  centre  of  ADE :  the  straight  line  which  joins  the  points 
F,  G  shall  pass  through  the  point  of  contact  A. 

Far,  if  not,  let  it  pass  otherwise,  if  possible,  as  FCDG,  and 


78 


THE  ELEMENTS 


Book  III.  join  FA,  AG:  and  because  F  is  the  centre  of  the  circle  ABC, 
*— Y'"-.'  AF  is  equal  to  FC  :  also,  ^  E 

because  G  is  the  centre  of 

the  circle  ADE,  AG  is  e- 

qual     to    GD :    therefore 

FA,  AG  are  equal  to  FC, 

DG ;  wherefore  the  whole 

FG  is    greater  than    FA, 
a  20. 1.     AG:    but  it  is  also  less*; 

which       is      impossible  : 

therefore  the  straight  line  which  joins  the  points  F,  G  shall  not 

pass  otherwise  than  through  the  point  of  contact  A,  that  is,  it 

must  pass  through  it.     Therefore,  if  two  circles,  Sec.     Q.  E.  D. 


PROP.  XIII.    THEOR. 

See  N.  ONE  circle  cannot  touch  another  in  more  points 
than  one,  whether  it  touches  it  on  the  inside  or 
outside. 

For,  if  it  be  possible,  let  the  circle  EBF  touch  the  circle  ABC 

in  more  points  than  one,  and  first  on  the  inside,  in  the  points 

a  10111.  B,  D;  join  BD,  and  draw*  GH  bisecting  BD  at  right  angles: 

Therefore,  because  the  points  B,  D  are  in  the  circumference  of 


b  2.  3.  "  each  of  the  circles,  the  straight  line  BD  falls  within  each**  of 
cCor.  1.3.  them  :  and  their  centres  are«=  in  the  straight  line  GH  which  bi- 
sects BD  at  right  angles;  therefore  GH  passes  through  the  point 
d  11.  3.  of  contact  J  ;  but  it  does  not  pass  through  it,  because  the  points 
li,  D  are  without  the  straight  line  GH,  which  is  absurd:  there- 
fore one  circle  cannot  touch  another  on  the  inside  in  more  points 
than  one. 


OF  EUCLID. 


79 


Nor  can  two  circles  touch  one  another  on  the  outside  in  Book  III. 
more  than  one  point:  for,  if  it  be  possible,  let  the  circle  ACK  <^-ymJ 
touch  the  circle  ABC  in  the  points  A,  C,  and  join  AC :  there- 
fore, because  the  two  points  A,  C  are  in 
the  circumference  of  the  circle  ACK,  the 
straight  line  AC  which  joins  them  shall 
fall  within *»  the  circle  ACK:  and  the  cir- 
cle ACK  is  without  the  circle  ABC  ;  and 
therefore  the  straight  line  AC  is  without 
this  last  circle  ;  but,  because  the  points  A, 
C  are  in  the  circumference  of  the  circle 
ABC,  the  straight  line  AC  must  be  within^* 
the  same  circle,  which  is  absurd :  there- 
fore one  circle  cannot  touch  another  on 
the  outside  in  more  than  one  point :  and  it 
has  been  shown,  that  they  cannot  touch  on 
the  inside  in  more  points  than  one.  There- 
fore, one  circle,  &c.     Q.  E.  D. 


PROP.  XIV.    THEOR. 


EQUAL  straight  lines  in  a  circle  are  equally  dis- 
tant  from  the  centre;  and  those  which  are  equally 
distant  from  the  centre,  are  equal  to  one  another. 


Let  the  straight  lines  AB,  CD,  in  the  circle  ABDC,  be  equal 
to  one  another ;  they  are  equally  distant  from  the  centre. 

Take  E  the  centre  of  the  circle  ABDC,  and  from  it  draw  EF, 
EG  perpendiculars  to  AB,  CD :  then,  because  the  straight  line 
EF,  passing  through  the  centre,  cuts  the  straight  line  AB,  which 
does  not  pass  through  the  centre,  at  right 
angles,  it  also  bisects  »  it :  wherefore  AF 
is  equal  toFB,  and  AB  double  of  AF.  For 
the  same  reason,  CD  is  double  of  CG  : 
and  AB  is  equal  to  CD  ;  therefore,  AF  is 
equal  to  CG:  and  because  AE  is  equal 
to  EC,  the  square  of  AE  is  equal  to  the 
square  of  EC  ;  but  the  squares  of  AF,  FE 
are  equal  b  to  the  square  of  AE,  because 
the  angle  AFE  is  a  right  angle ;  and,  for 
the  like  reason,  the  squares  of  EG,  GC  are  equal  to  the  square  of 
EC :  therefore  the  squares  of  AF,  FE  are  equal  to  the  squares  of 
vG,  GE,  of  which  the  square  of  AF  is  equal  to  the  square  of 


a  3. 


b  4r.  1. 


80 


THE  ELEMENTS 


Book  III.  CG,  because  AF  is  equal  to  CG  ;  therefore  the  remaining  square 

V— Y-^  of  FE  is  equal  to  the  remaining  square  of  EG,  and  the  straight 

line  EF  is  therefore  equal  to  EG :  but  straight  lines  in  a  circle 

are  said  to  be  equally  distant  from  the  centre,  when  the  perpen- 

c4.def.3.  diculars  drawn  to  them  from  the  centre  are  equal  <= :  therefoi'c 

AB,  CD  are  equally  distant  from  the  centre. 

Next,  if  the  straight  lines  AB,  CD  be  equally  distant  from 
the  centre,  that  is,  if  FE  be  equal  to  EG,  AB  is  equal  to  CD: 
for,  the  same  construction  being  made,  it  may,  as  before,  be  de- 
monstrated, that  AB  is  double  of  AF,  and  CD  double  of  CG, 
and  that  the  squares  of  EF,  FA  are  equal  to  the  squares  of  EG, 
GC ;  of  which  the  square  of  FE  is  equal  to  the  square  of  EG, 
because  FE  is  equal  to  EG  ;  therefore  the  remaining  square  of 
AF  is  equal  to  the  remaining  square  of  CG ;  and  the  straight 
line  AF  is  therefore  equal  to  CG :  and  AB  is  double  of  AF,  and 
CD  double  of  CG ;  wherefore  AB  is  equal  to  CD.  Therefore, 
equal  straight  lines,  &c.     Q.  E.  D. 


PROP.  XV.    THEOR. 


See  N  THE  diameter  is  the  greatest  straight  line  in  a 
circle;  and,  of  all  others,  that  which  is  nearer  to  the 
centre  is  always  greater  than  one  more  remote ;  and 
the  greater  is  nearer  to  the  centre  than  the  less. 


20.1. 


Let  ABCD  be  a  circle,  of  which  the 
diameter  is  AD,  and  the  centre  E;  and 
let  BC  be  nearer  to  the  centre  than  FG  ; 
AD  is  greater  than  any  straight  line  BC 
which  is  not  a  diameter,  and  BC  greater 
than  FG. 

From  the  centre  draw  EH,  EK  per- 
pendiculars to  BC,  FG,  and  join  EB, 
EC,  EF;  and  because  AE  is  equal  to 
EB,  and  ED  to  EC,  AD  is  equal  to  EB, 
EC  :  but  EB,  EC  are  greater  »  than  BC; 
wherefore  also  AD  is  greater  than  BC 

And,  because  BC  is  nearer  to  the   centre  than  FG,  EH  is 


OF  EUCLID.  81 

less**  than  EK ;  but,  as  was  demonstrated  in  the  preceding,  BC  Booklll. 
is  double  of  BH,  and  FG  double  of  FK,  and  the  squares  of  EH   ^■— v— ^ 
HB  are  equal  to  the  squares  of  EK,  KF,  of  which  the  square  ofb5.def.3. 
EH  is  less  than  the  square  of  EK,  because  EH  is  less  than  EK  ; 
therefore  the  square  of  BH  is  greater  than  the  square  of  FK, 
and  the  straight  line  BH  greater  than  FK  ;  and  therefore  BC  is 
greater  than  FG. 

Next,  Let  BC  be  greater  than  FG ;  BC  is  nearer  to  the  cen- 
tre than  FG,  that  is,  the  same  construction  being  made,  EH  is 
less  than  EK  :  because  BC  is  greater  than  FG,  BH  likewise  is 
greater  than  KF  :  and  the  squares  of  BH,  HE  are  equal  to  the 
squares  of  FK,  KE,  of  which  the  square  of  BH  is  greater  than 
the  square  of  FK,  because  BH  is  greater  than  FK  ;  therefore 
the  square  of  EH  is  less  than  the  square  of  EK,  and  the  straight 
line  EH  less  than  EK.     Wherefore  the  diameter,  &c.    Q.  E.  D. 


PROP.  XVL    THEOR. 

THE  straight  line  drawn  at  right  angles  to  the  dia-  See  n. 
meter  of  a  circle,  from  the  extremity  of  it,  falls  with- 
out the  circle ;  and  no  straight  line  can  be  drawn 
between  that  straight  line  and  the  circumference  from, 
the  extremity,  so  as  not  to  cut  the  circle ;  or,  which 
is  the  same  thing,  no  straight  line  can  make  so  great 
an  acute  angle  with  the  diameter  at  its  extremity,  or 
so  small  an  angle  with  the  straight  line  which  is  at 
right  angles  to  it,  as  not  to  cut  the  circle. 

Let  ABC  be  a  circle,  the  centre  of  which  is  D,  and  the  dia- 
meter AB  ;  the  straight  line  drawn  at  right  angles  to  AB  from 
its  extremity  A,  shall  fall  without  the  circle. 

For,  if  it  does  not,  let  it  fall,  if 
possible,  within  the  circle,  as  AC, 
and  draw  DC  to  the  point  C  where 
it  meets  the  circumference  :  and 
because  DA  is  equal  to  DC,  the 

angle  DAC  is  equal*  to  the  angle  (  y^  ~)  "^       *  ^-  '^ 

ACD  ;  but  DAC  is  a  right  angle, 
therefore  ACD  is  a  right  angle, 
and  the  angles  DAC,  ACD  are 
therefore  equal  to  two  right  an- 
gJesj  which  is  impossible'':  therefor^  the  straight  line  drawn b  17. 1. 

L 


82  THE  ELEMENTS 

Book  III.  from  A  at  right  angles  to  BA  does  not  fall  within  the  circle  :  in 

*— (T-*-'  the  same  manner  it  may  be  demonstrated,  that  it  does  not  fall 
upon  the  circumference  ;  therefore  it  must  fall  without  the  cir- 
cle, as  AE. 

And  between  the  straight  line  AE  and  the  circumference  no 
straight  line  can  be  drawn  from  the  point  A  which  does  not 
cut  the  circle  :  for,  if  possible,  let  FA   be  between  them,  and 

cl2.  1.  from  the  point  D  draw^  DG  perpendicular  to  FA,  and  let  it 
meet   the   circumference  in  H  :  and  because  AGD   is  a  right 

d  19. 1.  angle,  and  EAG  less  ^  than  a  right  angle  :  DA  is  greater  ^  than 
DG:    but    DA    is    equal   to   DH ;  " 

therefore  DH  is  greater  than  DG, 
the  less  than  the  greater,  which  is 
impossible :  therefore  no  straight 
line  can  be  drawn  from  the  point 
A  between  AE  and  the  circumfe- 
rence, which  does  not  cut  the  cir- 
cle, or,  which  amounts  to  the  same 
thing,  however  great  an  acute  angle 
a  straight  line  makes  with  the  dia- 
meter at  the  point  A,  or  however 
small  an  angle  it  makes  with  AE, 
the  circumference  passes  between 
that  straight  line  and  the  perpendicular  AE.  '  And  this  is  •all 
'  that  is  to  be  understood,  when,  in  the  Greek  text  and  transla- 
'  tions  from  it,  the  angle  of  the  semicircle  is  said  to  be  greater 
'  than  any  acute  rectilineal  angle,  and  the  remaining  angle  less 
'  than  any  rectilineal  angle.' 

Cor.  From  this  it  is  manifest,  that  the  straight  line  which  is 
drawn  at  right  angles  to  the  diameter  of  a  circle  from  the  ex- 
tremity of  it,  touches  the  circle  ;  and  that  it  touches  it  only  in 
one  point,  because,   if  it  did  meet  the  circle  in  two,  it  would 

e  2.  3.  fall  within  it  <=.  '  Also  it  is  evident  that  there  can  be  but  one 
*  straight  line  which  touches  the  circle  in  the  same  point.' 


PROP.  XVn.    PROB. 

TO  draw  a  straight  line  from  a  given  point,  either 
without  or  in  the  circumference,  which  shall  touch  a 
given  circle. 

First,  Let  A  be  a  given  point  without  the  given  circle  BCD  ; 


OF  EUCLID. 


it  is  required  to  draw  a  straight  line  from  A  whicli  shall  touch  Book  III. 
the  circle.  ^— -v"*.^ 

Find  a  the  centre  E  of  the  circle,  and  join  AE  ;  and  from  the  a  1.  3. 
centre  E,  at  the  distance  EA,  describe  the  circle  AFG  ;   from 
the  point  D  draw  ^  DF  at  right  angles  to  EA,  and  join  EBF,  AB.  b  11. 1. 
AB  touches  the  circle  BCD. 

Because     E    is    the    centre 
of   the    circles     BCD,     AFG, 
EA  is  equal  to  EF :  and  ED  to 
EB  ;    therefore  the   two  sides 
AE,  EB  are  equal  to  the  two 
FE,  ED,  and  they  contain  the 
angle  at  E  common  to  the  two 
triangles  AEB,   FED ;    there- 
fore the  base  DF  is  equal  to 
the    base  AB,    and   the   trian- 
gle FED  to  the  triangle  AEB, 
and  the  other  angles  to  the  other  angles  «  :  therefore  the  angle  c  4. 1. 
EBA  is  equal  to  the  angle  EDF  :  but  EDF  is  a  right  angle, 
wherefore  EBA  is  a  right  angle  :    and  EB  is  drawn  from  the    *»> 
centre  :    but  a   straight  line  di'awn   from  the  extremity  of  a 
diameter,  at  right  angles  to  it,  touches  the  circle^:  therefore d Cor.  16 
AB  touches  the  circle  ;  and  it  is  drawn  from  the  given  point  A.   3. 
Which  was  to  be  done. 

But,  if  the  given  point  be  in  the  circumference  of  the  circle, 
as  the  point  D,  draw  DE  to  the  centre  E,  and  DF  at  right  angles 
to  DE  ;  DF  touches  the  circle  •*. 


PROP.  XVIII.    THEOR. 


IF  a  Straight  line  touches  a  circle,  the  straight  line 
drawn  from  the  centre  to  the  point  of  contact  shall 
be  perpendicular  to  the  line  touching  the  circle. 

Let  the  straight  line  DE  touch  the  circle  ABC  in  the  point 
C  ;  take  the  centre  F,  and  draw  the  straight  line  FC  :  FC  is  per- 
pendicular to  DE. 

For,  if  it  be  not,  from  the  point  F  draw  FBG  perpendicular 
to  DE  ;  and  because  FGC  is  a  right  angle,  GCF  is  a  an  acute  a  17. 1 
angle  ;  and  to  the  greater  angle  the  greatest  ^  side  is  opposite  :  b  19. 1 


84 


THE  ELEMENTS 


Book  III.  therefore  FC  is  greater  than   FG  ; 

v.i-v-*^  but  FC  is  equal  to  FB  ;  therefore 
FB  is  greater  than  FG,  the  less 
than  the  greater,  which  is  impos- 
sible ;  wherefore  FG  is  not  per- 
pendicular to  DE :  in  the  same 
manner  it  may  be  shown,  that  no 
other  is  perpendicular  to  it  besides 
FC,  that  is,  FC  is  perpendicular  to 
DE.  Therefore,  if  a  straight  line, 
&c.    Q.  E.  D. 


a  18.  3. 


PROP.  XIX.     THEOR. 

IF  a  straight  line  touches  a  circle,  and  from  the 
point  of  contact  a  straight  line  be  drawn  at  right  an- 
gles to  the  touching  line,  the  centre  of  the  circle  shall 
be  in  that  line. 

Let  the  straight  line  DE  touch  the  circle  ABC  in  C,  and  from 
C  let  CA  be  drawn  at  right  angles  to  DE  ;  the  centre  of  the  cir- 
cle is  in  CA. 

For,  if  not,  let  F  be  the  centre,  if  possible,  and  join  CF : 
because    DE    touches    the    circle  A 

ABC,  and  FC  is  drawn  from  the 
centre  to  the  point  of  contact,  FC 
is  perpendicular  »  to  DE  ;  there- 
fore FCE  is  a  right  angle:  but  ACE 
is  also  a  right  angle  ;  therefore 
the  angle  FCE  is  equal  to  the  an- 
gle ACE,  the  less  to  the  greater, 
which  is  impossible  :  wherefore  F 
is  not  the  centre  of  the  circle  ABC: 
in  the  same  njanner  it  may  be 
shown  that  no  other  point  which  is 
not  in  CA  is  the  centre  ;  that  is, 


D 


C 


the  centre  is  in  CA.     There- 


fore, if  a  straight  line,  Sec.     Q.  E.  D, 


PROP.  XX.     THEOR. 


See  N.  THE  angle  at  the  centre  of  a  circle  is  double  of 
the  angle  at  the  circumference,  upon  the  same  base, 
that  is',  upon  the  same  part  of  the  circumference. 


OF  EUCLID. 


b32. 1 


Let  ABC  be  a  circle,  and  BEC  an  angle  at  the  centre,  and  Book  III, 
BAC  an  angle  at  the  circumference,  which  have  the  same  cir-  <^v^ 
cumference  BC  for  their  base  ;  the  angle 
BEC  is  double  of  the  angle  BAC. 

First,  let  E  the  centre  of  the  circle  be 
within  the  angle  BAC,  and  join  AE,  and 
produce  it  to  F :  because  EA  is  equal  to 
EB,  the  angle  EAB  is  equal  ^  to  the 
angle  EBA ;  therefore  the  angles  EAB, 
EBA  are  double  of  the  angle  EAB ;  but 
the  angle  BEF  is  equal ''  to  the  angles 
EAB,  EBA ;  therefore  also  the  angle  BEF 
is  double  of  the  angle  EAB :  for  the  same 
reason,  the  angle  FEC  is  double  of  the 

angle  EAC :  therefore  the  whole  angle  BEC  is  double  of  the 
whole  angle  BAC. 

Again,  Let  E  the  centre  of  the  A 

circle  be  without  the  angle  BDC,  and 
join  DE,  and  produce  it  to  G.  It 
may  be  demonstrated,  as  in  the  first 
case,  that  the  angle  GEC  is  double 
of  the  angle  GDC,  and  that  GEB  a 
part  of  the  first  is  double  of  GDB  a 
part  of  the  other ;  therefore  the  re- 
maining angle  BEC  is  double  of  the 
remaining  angle  BDCr  Therefore, 
the  angle  at  the  centre,  8cc,  Q.  E.  D. 


PROP.  XXL    THEOR. 


THE  angles  in  the  same  segment  of  a  circle  are  Sec  n 
equal  to  one  another. 


Let  ABCD  be  a  circle,  and  BAD, 
BED  angles  in  the  same  segment 
BAED :  the  angles  BAD,  BED  are 
equal  to  one  another. 

Take  F  the  centre  of  the  circle 
ABCD:  and,  first,  let  the  segment 
BAED  be  greater  than  a  semicircle, 
and  join  BF,  FD  :  and  because  the 
angle  BED  is  at  the  centre,  and  the 
angle  BAD  at  the  circumference, 
and  that  they  have  the  same  part  of 


86 


THE  ELEMENTS 


Booklll.  the  circumference,  viz.  BCD,  for  their  base ;  therefore  the  an- 
*>— -v— ^  gle  BFB  is  double  »  of  the  angle  BAD :  for  the  same  reason, 
a  20. 3.     the  angle  BFD  is  double  of  the  angle  BED:  therefore  the  angle 

BAD  is  equal  to  the  angle  BED. 

But,  if  the  segment  BAED  be  not  greater  than  a  semicircle, 

let  BAD,  BED  be  angles  in  it;  these 

also  are  equal  to  one  another:  draw 

AF  to  the  centre,  and  produce  it  to 

C,  and  join  CE  :  therefore  the  seg- 
ment BADC  is  greater  than  a  semi- 
circle ;    and   the  angles    in    it   BAG, 

BEC  are  equal,  by  the  first  case :  for 

the  same   reason,  because  CBED   is 

greater  than  a  semicircle,  the  angles 

CAD,  CED  are  equal :   therefore  the 

whole    angle    BAD    is    equal    to    the 

whole    angle  BED.     Wherefore,  the 

angles  in  the  same  segment,  &:c.     Q.  E.  D. 


PROP.  XXII.    THEOR. 


THE  opposite  angles  of  any  quadrilateral  figure  de- 
scribed in  a  circle  are  together  equal  to  two  right 


angles. 


Let  ABCD  be  a  quadrilateral  figure  in  the  circle  ABCD ;  any 

two  of  its  opposite  angles  are  together  equal  to  two  right  angles. 

Join  AC,  BD  ;   and    because  the    three  angles    of  every   tri- 

a  32. 1.     angle  are  equal*  to  two  right  angles,   the  three  angles  of  the 

triangle  CAB,  viz.  the  angles  CAB,  ABC,  BCA  are  equal  to 

two  right  angles :  but  tlic  angle  CAB 
b21.  3.     is  equal''  to  the  angle  CDB,  because 

they  are  in  the  same  segment  BADC, 

and  the   angle  ACB   is   equal   to   the 

angle  ADB,  because  they  are  in  the 

same  segment  ADCB:  therefore  the 

whole    angle    ADC    is  equal    to    the 

angles  CAB,   ACB :  to  each  of  these 

equals  add  the  angle  ABC  ;  therefore 

the    angles    ABC,    CAB,    BCA    are 

equal  to  the  angles  AHC,  ADC :  but  7\BC,  CAB,  BCA    are 

equal  to  two  right  angles;  therefore  also  the  angles  ABC,  ADC 

are  equal  to  tv.o  right  angles:  in  the  same  manner,  the  angles 


OF  EUCLID;  87 

BAD,  DCB  may  be  shown  to  be  equal  to  two  right  angles.  Book  III. 
Therefore,  the  opposite  angles,  £cc.     Q.  E.  D.  •— v— ' 


PROP.  XXIII.   THEOR. 

UPON  the  same  straight  line,  and  upon  the  same  See  n. 
side  of  it,  there  cannot  be  two  similar  segments  of 
circles  not  coinciding  with  one  another. 

If  it  be  possible,  let  the  two  similar  segm-ents  of  circles,  viz.  ■ 
ACB,  ADB,  be  upon  the  same  side  of  the  same  straight  line 
AB,  not  coinciding   with  one   another :    then,  because  the  cir- 
cle ACB  cuts  the  circle  ADB  in  the  two 
points  A,  B,  they  cannot  cut  one  another  ^ 

in  any  other  point  ^:    one  of  the  segments  /^X    \('^-^      ^  ^^'  ^' 

must  therefore  fall  within  the  other;    let 
ACB  fall  within  ADB,  and  draw  the  straight 
line  BCD,  and  join  C A,  DA :  and  because 
the  segment  ACB  is  similar  to  the  segment 
ADB,  and  that  similar  segments  of  circles  contain  ^  equal  an-  b  11.  def. 
gles  ;  the  angle  ACB  is  equal  to  the  angle  ADB,  the  exterior  to     3. 
the  interior,  which  is  impossible  <=.     Therefore,  there  cannot  be  c  16. 1. 
two  similar  segments  of  a  circle  upon  the  same  side  of  the  same 
line,  which  do  not  coincide.     Q.  E.  D. 


PROP.  XXIV.    THEOR. 

SIMILAR  segments  of  circles  upon  equal  straight  See  n 
lines  are  equal  to  one  another. 

Let  AEB,  CFD  be  similar  segments  of  circles  upon  the  equal 
straight  lines  AB,  CD ;  the  segment  AEB  is  equal  to  the  seg- 
ment CFD. 

For,  if  the  seg~  E  F 

ment  AEB  be  ap- 
plied to  the  seg- 
ment CFD,  so  as 
the  point  A  be  on 
C,  and  the  straight 
line  AB  upon  CD,  the  point  B  shall  coincide  with  the  point  D, 


88 


THE  ELEMENTS 


Book  III.  because  AB  is  equal  to  CD  :  therefore,  the  straight  line  AB  co- 
<^^^fmmJ  inciding  with  CD,  the  segment  AEB  must  *  coincide  with  the 
%  23.  3.    segment  CFD,  and  therefore  is  equal  to  it.     Wherefore,  similar 
segments,  &c.    Q.  E.  D. 


PROP.  XXV.    PROB. 

See  N.        A  SEGMENT  of  a  circle  being  given,  to  describe 
the  circle  of  which  it  is  the  segment. 


a  10.  1. 
b  11.  1. 
c6. 1. 


d9.3. 


Let  ABC  be  the  given  segment  of  a  circle ;  it  is  required  to 
describe  the  circle  of  which  it  is  the  segment. 

Bisect  ^  AC  in  D,  and  from  the  point  D  draw  ^  DB  at  right 
angles  to  AC,  and  join  AB:  first,  let  the  angles  ABD,  BAD 
be  equal  to  one  another;  then  the  straight  line  BD  is  equal '^ 
to  DA,  and  therefore  to  DC ;  and  because  the  three  straight 
lines  DA,  DB,  DC  are  all  equal,  D  is  the  centre  of  the  cir- 
cle *!:  from  the  centre  D,  at  the  distance  of  any  of  the  three 
DA,  DB,  DC  describe  a  circle  ;  this  shall  pass  through  the  other 
points ;  and  the  circle  of  which  ABC  is  a  segment  is  described : 
and  because  the  centre  D  is  in  AC,  the  segment  ABC  is  a  se- 


micircle: but  if  the  angles  ABD,  BAD  are  not  equal  to  one 
e  23. 1.  another,  at  the  point  A,  in  the  straight  line  AB,  make  «  the  angle 
BAE  equal  to  the  angle  ABD,  and  produce  BD,  if  necessary,  to 
E,  and  join  EC :  and  because  the  angle  ABE  is  equal  to  the  an- 
gle BAE,  the  straight  line  BE  is  equal  e  to  EA  :  and  because  AD 
is  equal  to  DC,  and  DE  common  to  the  triangles  ADE,  CDE, 
the  two  sides  AD,  DE  are  equal  to  the  two  CD,  DE,  each  to 
each  ;  and  the  angle  ADE  is  equal  to  the  angle  CDE,  for 
each  of  them  is  a  right  angle;  therefore  the  base  AE  is  equal 
f  4. 1.  f  to  the  base  EC  :  but  AE  was  shown  to  be  equal  to  EB  ;  where- 
fore also  BE  is  equal  to  EC:    and  the  three  straight  lines  AE; 


OF  EUCLID. 


89 


EB,  EC  are  therefore  equal  to  one  another;  Avherefore*  E  is  Book  III. 
the  centre  of  the  circle.     From  the  centre  E,  at  the  distance  of  ^  -y^ 
any  of  the  three  AE,  EB,  EC,  describe  a  circle ;  this  shall  pass  d  9.  3. 
through  the  other  points ;  and  the  circle  of  which  ABC  is  a  seg- 
ment is  described:  and  it  is  evident,  that  if  the  angle  ABD  be 
greater  than  the  angle  BAD,  the  centre  E  falls  without  the  seg- 
ment ABC,  which  therefore  is  less  than  a  semicircle :  but  if  the 
angle  ABD  be  less  than  BAD,  the  centre  E  falls  within  the  seg- 
ment ABC,  which  is  therefore  greater  than  a  semicircle  :  where- 
fore a  segment  of  a  circle  being  given,  the  circle  is  described  of 
which  it  is  a  segment.     Which  was  to  be  done. 


PROP.  XXVI.   THEOR. 

IN  equal  circles,  equal  angles  stand  upon  equal 
circumferences,  whether  they  be  at  the  centres  or 
circumferences. 


Let  ABC,  DEF  be  equal  circles,  and  the  equal  angles  BGC, 
EHF  at  their  centres,  and  BAC,  EDF  at  their  circumferences  : 
the  circumference  BKC  is  equal  to  the  circumference  ELF. 

Join  BC,  EF ;  and  because  the  circles  ABC,  DEF  are  equal, 
the  straight  lines  drawn  from  their  centres  are  equal:  there- 
fore the  two  sides  BG,  GC  are  equal  to  the  two  EH,  HF; 


and  the  angle  at  G  is  equal  to  the  angle  at  H ;  therefore  the  base 
BC  is  equal  ^  to  the  base  EF  :  and  because  the  angle  at  A  is  equal  a  4. 1. 
to  the  angle  at  D,  the  segment  BAC  is  similar  •>  to  the  segment  b  11.  def. 
EDF ;  and  they  are  upon  equal  straight  lines  BC,  EF ;  but  simi-   3. 
lar  segments  of  circles  upon  equal  straight  lines  are  equal  <=  to  * 

one  another :  therefore  the  segment  BAC  is  equal  to  the  segment  £.  94  " 
EDF  :  but  the  whole  circle  ABC  is  equal  to  the  whole  DEF ;        "  "^ 

M 


90 


THE  ELEMENTS 


Booklll.  therefore -the  remaining  segment  BKC  is  equal  to  the  remaining 
^^yi^  segment  ELF,  and  the  circumference  BKC  to  the  circumference 
ELF.     Wherefore,  in  equal  circles,  &;c.     Q.  E.  D. 


PROP.  XXVn.    THEOR. 

IN  equal  circles,  the  angles  which  stand  upon  equal 
circumferences  are  equal  to  one  another,  whether  they 
be  at  the  centres  or  circumferences. 


a  20.  3. 


Let  the  angles  BGC,  EHF  at  the  centres,  and  BAG,  EDF  at 
the  circumferences  of  the  equal  circles  ABC,  DEF  stand  upon 
the  equal  circumferences  BC,  EF :  the  angle  BGC  is  equal  to 
the  angle  EHF,  and  the  angle  BAG  to  the  angle  EDF. 

If  the  angle  BGC  be  equal  to  the  'angle  EHF,  it  is  manifest  * 
that  the  angle  BAG  is  also  equal  to  EDF :  but,  if  not,  one  of 


them  is  the  greater :  let  BGC  be  the  greater,  and  at  the  point 
b  23. 1.  G,  in  the  straight  line  BG,  make^  the  angle  BGK  equal  to  the 
c26.  3.  angle  EHF;  but  equal  angles  stand  upon  equal  circumferences  S 
when  they  are  at  the  centre ;  therefore  the  circumference  BK  is 
equal  to  the  circumference  EF :  but  EF  is  equal  to  BC ;  there- 
fore also  BK  is  equal  to  BC,  the  less  to  the  greater,  which  is 
impossible :  therefore  the  angle  BGC  is  not  unequal  to  the  an- 
gle EHF;  that  is,  it  is  equal  to  il :  and  the  angle  at  A  is  half  of 
the  angle  BGC,  and  the  angle  at  D  half  of  the  angle  EHF  :  there- 
fore the  angle  at  A  is  equal  to  the  angle  at  D.  Wherefore,  in 
equal  circles,  S^c     Q-  E.  D. 


OF  EUCLID. 


PROP.  XXVIII.    THEOR. 

IN  equal  circles,  equal  straight  lines  cut  off  equal 
circumferences,  the  greater  equal  to  the  greater,  and 
the  less  to  the  less. 

Let  ABC,  DEF  be  equal  circles,  and  BC,  EF  equal  straight 
lines  in  them,  which  cut  off  the  two  greater  circumferences 
BAC,  EDF,  and  the  two  less  BGC,  EHF;  the  greater  BAC 
is  equal  to  the  greater  EDF,  and  the  less  BGC  to  the  less  EHF. 

Take  »  K,  L  the  centi-es  of  the  circles,  and  join  BK,  KC,  a- 1-  3- 
EL,  LF :   and  because  the  circles  are  equal,  the  straight  lines 


from  their  centres  are  equal;  therefore  BK,  KC  are  equal  to 
EL,  LF ;  and  the  base  BC  is  equal  to  the  base  EF ;  therefore 
the  angle  BKC  is  equal  ^  to  the  angle  ELF :  but  equal  angles  b  8. 1. 
stand  upon  equal  <=  circumferences,  when  they  are  at  the  cen-c26. 3. 
tres ;  therefore  the  circumference  BGC  is  equal  to  the  circum- 
ference EHF.  But  the  whole  circle  ABC  is  equal  to  the  whole 
EDF ;  the  remaining  part,  therefore,  of  the  circumference,  viz. 
BAC,  is  equal  to  the  remaining  part  EDF.  Therefore,  in  equal 
circles,  &:c.     Q.  E.  D. 


PROP.  XXIX.    THEOR. 

IN  equal  circles,  equal  circumferences  are  subtended 
by  equal  straight  lines. 

Let  ABC,  DEF  be  equal  circles,  and  let  the  circumferences 
BGC,  EHF  also  be  equal ;  and  join  BC,  EF :  the  straight  line 
BC  is  equal  to  the  straight  line  EF. 


92 


THE  ELEMENTS 


Book  III.  Take  »  K,  L,  the  centres  of  the  circles,  and  join  BK,  KC, 
v^.i'V—^  EL,  LF :  and  because  the  circumference  BGC  is  equal  to  the 
a  1.  3. 

A  D 


G 


H 


b27. 3.  circumference  EHF,  the  angle  BKC  is  equal  ^  to  the  angle 
ELF :  and  because  the  circles  ABC,  DEF  are  equal,  the  straight 
lines  from  their  centres  are  equal :  therefore  BK,  KC  are  equal 
to  EL,  LF,  and  they  contain  equal  angles :    therefore  the  base 

c  4. 1.  BC  is  equal  <=  to  the  base  EF.  Therefore,  in  equal  circles,  8cc. 
Q.  E.  D. 


PROP.  XXX.    PROB. 


a  10.  1. 


TO  bisect  a  given  circumference,  that  is,  to  dhide 
it  into  two  equal  parts. 

Let  ADB  be  the  given  circumference  ;  it  is  required  to  bisect  it. 

Join  AB,  and  bisect  »  it  in  C ;  from  the  point  C  draw  CD  at 
right  angles  to  AB,  and  join  AD,  DB :  the  circumference  ADB 
is  bisected  in  the  point  D. 

Because  AC  is  equal  to  CB,  and  CD  common  to  the  triangles 
ACD,  BCD,  the  two  sides  AC,  CD 
are  equal  to  the  two  BC,  CD ;  and 
the  angle  ACD  is  equal  to  the  angle 
BCD,  because  each  of  them  is  a  right 
angle  ;  therefore  the  base  AD  is  equal 
^  to  the  base  BD :    but  equal  straight 

lines  cut  off  equal  <=  circumferences,  the  greater  equal  to  the 
greater,  and  the  less  to  the  less,  and  AD,  DB  are  each  of  them 
d  Cor,  1.  less  than  a  semicircle  ;  because  DC  passes  through  the  centre  ^  : 
'•  wherefore  the  circumference  AD  is  equal  to  the  circumference 
DB :  therefore  the  given  circumference  is  bisected  in  D.  Which 
was  to  be  done. 


b4.  1. 

c28.3 


OF  EUCLID. 


PROP.  XXXI.    THEOR. 


IN  a  circle,  the  angle  in  a  semicircle  is  a  right  angle ; 
but  the  angle  in  a  segment  greater  than  a  semicircle 
is  less  than  a  right  angle ;  and  the  angle  in  a  segment 
less  than  a  semicircle  is  greater  than  a  right  angle. 

Let  ABCD  be  a  circle,  of  which  the  diameter  is  BC,  and 
centre  E;  and  draw  CA  dividing  the  circle  into  the  segments 
ABC,  ADC,  and  join  BA,  AD,  DC  ;  the  angle  in  the  semi- 
circle BAG  is  a  right  angle ;  and  the  angle  in  the  segment 
ABC,  which  is  greater  than  a  semicircle,  is  less  than  a  right 
angle ;  and  the  angle  in  the  segment  ADC,  which  is  less  than  a 
semicircle,  is  greater  than  a  right  angle. 

Join  AE,  and  produce  BA  to  F ;  and  because  BE  is  equal 
to  EA,  the  angle  EAB  is  equal  ^  to  EBA;  also,  because  AE  a_5. 1. 
is  equal  to  EC,  the  angle  EAC  is 
equal  to  ECA;  wherefore  the 
whole  angle  BAC  is  equal  to  the 
two  angles  ABC,  ACB  ;  but  FAC, 
the  exterior  angle  of  the  triangle 
ABC,  is  equal  ^  to  the  two  angles 
ABC,  ACB;  therefore  the  angle 
BAC  is  equal  to  the  angle  FAC, 
and  each  of  them  is  therefore  a 
rights  angle  :  wherefore  the  angle 
BAC  in  a  semicircle  is  a  right 
angle. 

And   because   the   two    angles 
ABC,  BAC  of  the  triangle  ABC  are  together  less<i  than  two  d  17. 11 
right  angles,  and  that  Bx\C  is  a  right  angle,  ABC  must  be  less 
than  a  right  angle ;  and  therefore  the  angle  in  a  segment  ABC 
greater  than  a  semicircle,  is  less  than  a  right  angle. 

And  because  ABCD  is  a  quadrilateral  figure  in  a  circle,  any 
two  of  its  opposite  angles  are  equal*  to  two  right  angles  ;  there-  e  22.  3. 
fore  the  angles  ABC,  ADC  are  equal  to  two  right  angles ;  and 
ABC  is  less  than  a  right  angle ;  wherefore  the  other  ADC  is 
greater  than  a  right  angle. 

Besides,  it  is  manifest,  that  the  circumference  of  the  greater 
segment  ABC  falls  without  the  right  angle  CAB,  but  the 
circumference  of  the  less  segment  ADC  falls  within  the  right 
angle   CAF.     '   And    this  is  all   that    is    meant,    when  in  the 


b  32. 1. 


c  10. 
dcf.  1. 


94  THE  ELEMENTS 

Book  III. '  Greek  text,  and  the  translations  from  it,  the  angle  of  the  greater 
'■"■"•V-— ^  '  segment  is  said  to  be  greater,  and  the  angle  of  the  less  segment 

'  is  said  to  be  less,  than  a  right  angle.' 

Cor.     From  this  it  is  manifest,  that  if  one  angle  of  a  triangle 

be  equal  to  the  other  two,  it  is  a  right  angle,  because  the  angle 

adjacent  to  it  is  equal  to  the;  same  two;  and  when  the  adjacent 

angles  are  equal,  they  are  right  angles. 


PROP.  XXXII.     THEOR* 

IF  a  straight  line  touches  a  circle,  and  from  the 
point  of  contact  a  straight  line  be  drawn  cutting  the 
circle,  the  angles  made  by  this  line  with  the  line  touch- 
ing the  circle,  shall  be  equal  to  the  angles  which  are 
in  the  alternate  segments  of  the  circle. 

Let  the  straight  line  EF  touch  the  circle  ABCD  in  B,  and  from 
the  point  B  let  the  straight  line  BD  be  draAvn,  cutting  the  circle  : 
the  angles  which  BD  makes  with  the  touching  line  EF  shall  be 
equal  to  the  angles  in  the  alternate  segments  of  the  circle :  that 
is,  the  angle  FBD  is  equal  to  the  angle  which  is  in  the  segment 
DAB,  and  the  angle  DBE  to  the  angle  in  the  segment  BCD.  * 

a  11.  1.  From  the  point  B  draw  ^  BA  at  right  angles  to  EF,  and  take 
any  point  C  in  the  circumference  BD,  and  join  AD,  DC,  CB; 
and  because  the  straight  line  EF  touches  the  circle  ABCD  in 
the  point  B,  and  BA  is  drawn  at  ^ 

right  angles  to  the  touching  line 
from  the  point  of  contact  B,  the 

b  19.  3.     centre  of  the  circle  is  ^  in  BA  ; 
therefore    the  angle  ADB   in  a 

c  31.  3.     semicircle  is  a  rights  angle,  and 
consequently  the  other  two  angles 

A  32.  h  BAD,  ABD  are  equal  d  to  a  right 
angle:  but  ABF  is  likewise  a 
right  angle  ;  therefore  the  angle 
ABF  is  equal  to  the  angles  BAD, 
ABD  :  take  from  these  equals  the 
common  angle  ABD  ;  therefore  the  remaining  angle  DBF  is  equal 
to  the  angle  BAD,  which  is  in  the  alternate  segment  of  the  circle  ; 
and  because  ABCD  is  a  quadrilateral  figure  in  a  circle,  the  oppo- 

e  22  S.     site  angles  BAD,  BCD  are  equal «  to  two  right  angles  ;  therefore 


OF  EUCLID. 


93 


the  angles  DBF,  DBE,  being  likewise  equal*"  to  two  right  angles,  Book  HI. 
are  equal  to  the  angles  BAD,  BCD  ;  and  DBF  has  been  proved  *— v— ' 
equal  to  BAD  :  therefore  the  remaining  angle  DBE  is  equal  to^^^.  1. 
the  angle  BCD  in  the  alternate  segment  of  the  circle.     Where- 
fore, if  a  straight  line,  Stc.     Q.  E.  D. 


PROP.  XXXIII.     PROB. 


UPON  a  given  straight  line  to  describe  a  segment  ^^^  ^ 
of  a  circle,  containing  an  angle  equal  to  a  given  rec- 
tilineal angle. 

Let  AB  be  the  given  straight  line,  and  the  angle  at  C  the 
given  rectilineal  angle ;  it  is  required  to  describe  upon  the  given 
straight  line  AB  a  segment  of  a  circle,  containing  an  angle  equal 
to  the  angle  C. 

First,  Let  the  angle  at  C  be  a 
right  angle,  and  bisect »  AB  in  F, 
and  from  the  centre  F,  at  the  dis- 
tance FB,  describe  the  semicir- 
cle AHB  ;  therefore  the  angle 
AHB  in  a  semicircle  is  ^  equal 
to  the  right  angle  at  C 


10. 1, 


i^  b  31.  3. 


But,  if  the  angle  C  be  not  a  right  angle,    at  the  point  A,  in 
the  straight  line  AB,  make  <=  the  angle  BAD  equal  to  the  an-  c  23. 1. 
gle  C,  and  from  the  point  A 
draw  «i  AE  at  right  angles  to 
AD;  bisect  »  AB   in    F,   and 

from  F  draw  d  FG  at  right  /  J\     d  U-  U 

angles  to  AB,   and  join  GB: 
and   because   AF   is  equal  to 
FB,  and  FG  common  to  the 
triangles     AFG,    BFG,     the 
two  sides  AF,  FG  are  equal 
to  the  two  BF,  FG ;    and  the 
angle   AFG   is   equal  to  the 
angle    BFG ;     therefore    the 
base  AG  is  equal  «  to  the  base  GB  ;  and  the  circle  described  e  4.  L 
from  the  centre  G,  at  the  distance  GA,  shall  pass  through  the 
point  B  ;  ■  let  this  be  the  circle   AHB  :    and  because  from  the 
point  A,  the  extremity  of  the  diameter  AE,  AD  is  drawn  at 


96 


THE  ELEMENTS 


Book  III.  right  angles  to  AE,  therefore  AD  f  touches  the  circle;  and  be« 
■  cause  AB  drawn  from  the  point    q  ^ -.^.^^   u 

of  contact  A  cuts  the  circle,  the 

angle    DAB     is    equal    to    the 

angle  in  the  alternate  segment 
g  32.  3.     AHB  g  :    but  the  angle  DAB  is 

equal  to  the  angle  C,  therefore 

also  the  angle  C  is  equal  to  the 

angle    in    the    segment    AHB : 

wherefore,      upon      the     given 

straight  line    AB   the   segment 

AHB  of  a  circle  is  described,  which  contains  an  angle  equal  to 

the  given  angle  at  C.     Which  was  to  be  done. 


a  17.  3. 


b  'Jo.  1. 


PROP.  XXXIV.     PROB. 

TO  cut  off  a  segment  from  a  given  circle  which 
shall  contain  an  angle  equal  to  a  given  rectilineal 
angle. 

Let  ABC  be  the  given  circle,  and  D  the  given  rectilineal  angle  ; 
it  is  required  to  cut  off  a  segment  from  the  circle  ABC  that  shall 
contain  an  angle  equal  to  the  given  angle  D. 

Draw  »  the  straight  line  EF  touching  the  circle  ABC  in  the 
point  B,  and  at  the  point  B, 
in  the  straight  line  BF, 
make  '^  the  angle  FBC  equal 
to  the  angle  D  ;  therefore, 
because  the  straight  line 
EF  touches  the  circle  ABC, 
and  BC  is  drawn  from  the 
point  of  contact  B,  the  angle 
FBC  is  equal  =  to  the  angle 
in  the  alternate  segment 
BAC  of  the  circle  :  but  the 

angle  FBC  is  equal  to  the  angle  D  ;  therefore  the  angle  in  the  seg- 
ment BAC  is  equal  to  the  angle  D  :  Avherefore  the  segment  BAC 
is  cutoff  from  the  g'ven  circle  ABC  containing  an  angle  equal  to 
the  given  angle  D.     Which  was  to  be  done. 


OF  EUCLID. 


PROP.  XXXV.    THEOR. 


If  two  Straight  lines  within  a  circle  cut  one  ano-  See  n. 
ther,  the  rectangle  contained  by  the  segments  of  one 
of  them  is  equal  to  the  rectangle  contained  by  the 
segments  of  the  other. 


Let  the  two  straight  lines  AC,  BD,  within  the  circle  ABCD, 
cut  one  another  in  the  point  E  :  the  rectangle  contained  by  AE, 
EC  is  equal  to  the  rectangle  contained  by 
BE,  ED. 

If  AC,  BD  pass  each  of  them  through 
the  centre,  so  that  E  is  the  centre ;  it  is 
evident,  that  AE,  EC,  BE,  ED,  being  all 
equal,  the  rectangle  AE,  EC  is  likewise 
equal  to  the  rectangle  BE,  ED. 

But  let  one  of  them  BD  pass  through  the  centre,  and  cut  the 
other  AC,  which  does  not  pass  through  the  centre,  at  right  an- 
gles, in  the  point  E :  then,  if  BD  be  bisected  in  F,  F  is  the  cen- 
tre of  the  circle  ABCD  ;  join  AF :  and  because  BD,  which  passes 
through  the  centre,  cuts  the  straight  line  AC,  which  does  not 
pass  through  the  centre,  at  right  angles 
in  E,  AE,  EC  are  equal  ^  to  one  ano- 
ther :  and  because  the  straight  line  BD 
is  cut  into  two  equal  parts  in  the  point 
F,  and  into  two  unequal  in  the  point  E, 
the  rectangle  BE,  ED,  together  with 
the  square  of  EF,  is  equal  ^  to  the  square 
of  FB,  that  is,  to  the  square  of  FA ;  but  A 
the  squares  of  AE,  EF  are  equal  <=  to  the 
square  of  FA ;  therefore  the  rectangle 
BE,  ED,  together  with  the  square  of 
EF,  is  equal  to  the  squares  of  AE,  EF : 
lake  away  the  common  square  of  EF,  and  the  remaining  rectan- 
gle BE,  ED  is  equal  to  the  remaining  square  of  AE ;  that  is,  to 
the  rectangle  AE,  EC. 

Next,  Let  BD,  which  passes  through  the  centre,  cut  the 
other  AC,  which  does  not  pass  through  the  centre,  in  E,  but 
not  at  right  angles :  then,  as  before,  if  BD  be  bisected  in  F,  F 
Is  tfhe  centre  of  the  circle.     Join  AF,  and  from  F  draw  *  FG  d  12. 1. 

N 


a  3.  6. 


b5.2. 
c  47. 1 


98 


THE  ELEMENTS 


Book  III,  perpendicular  to  AC;  therefore  AG  is  equal*  to  GC ;  where- 
•>— v-^  fore  the  rectangle  AE,  EC,  together  with  the  square  of  EG,  is 
a  3.  5.       equal  ^  to  the  square  of  AG  :  to  each   of  these  equals  add  the 
hS.2.      square  of  GF;  therefore  the  rectangle  AE,  EC,  together  with 
the  squares  of  EG,  GF,  is  equal  to 
the    squares  of  AG,    GF  :    but   the 
c  47, 1.     squares  of  EG,  GF  are  equal  ^  to  the 
square  of  EF ;  and   the  squares    of 
AGj  GF  are  equal  to  the  square  of 
Af:     therefore   the    rectangle    AE, 
EC,  together  with  the  square  of  EF, 
is  equal  to  the  square  of  AF  ;    that 
is,    to  the    square    of  FB :    but  the 
square  of  FB  is  equal  "^  to  the  rectangle  BE,  ED,  together  with 
the  square  of  EF :   therefore  the  rectangle   AE,   EC,    together 
with  the  square  of  EF,  js  equal  to  the  rectangle  BE,  ED,  toge- 
ther with  the  square  of  EF :   take  away  the  common    square  of 
EF,  and  the  remaining  rectangle  AE,  EC  is  therefore  equal  to  the 
remaining  rectangle  BE,  ED. 

Lastly,  Let  neither  of  the  straight  lines  AC,  BD  pass  through 
the  centre :  take  the  centre  F,  and 
through  E,  the  intersection  of  the 
straight  lines  AC,  DB,  draw  the  dia- 
meter GEFH:  and  because  the  rect- 
angle AE,  EC  is  equal,  as  has  been 
shown,  to  the  rectangle  GE,  EH  ;  and, 
for  the  same  reason,  the  rectangle  BE, 
ED  is  equal  to  the  same  rectangle  GE, 
EH  ;  thficfore  the  rectangle  AE,  EC 
is  equal  to  the  rectangle  BE,  ED. 
Whers-fbre,  if  two  straight  lines,  £^c. 


PROP.  XXXVL    THEOR. 


IF  from  any  point  without  a  circle  two  straight 
lines  be  drawn,  one  of  which  cuts  the  circle,  and  the 
Other  touches  it;  the  rectangle  contained  by  the  whole 
Une  which  cuts  the  circle,  and  the  part  of  it  without 
the  circle,  shall  be  equal  to  the  square  of  the  li.n^ 
which  touches  it. 


OF  EUCLID* 


n 


c  47.  J. 


Let  D  be  any  point  without  the  circle  ABC,  and  I>CA,  DB  Book  III. 
two  straight  lines  drawn  from  it,  of  which  DCA  cuts  the  circle,  Vi^^— ^ 
and  DB  touches  the  same  :  the  rectangle  AD,  DC  is  equal  to  the 
square  of  DB. 

Either  DCA  passes  through  the  centre,  or  it  does  not ;  first, 
let  it  pass  through  the  centre  E,  and  join  EB  ;  therefore  the  an- 
gle EBD  is  a  right  *  angle  t  and  be-  a  18.  3» 
cause  the  straight  line  AC  is  bisected 
in  E,  and  produced  to  the  point  D,  the 
rectangle  AD,  DC,  together  with  the 

Square  of  EC,  is  equal  ^  to  the  square  /        C  b  6.  2. 

of  ED,  and  CE  is  equal  to  EB  :  there- 
fore the  rectangle  AD,  DC,  together 
with  the  square  of  EB,  is  equal  to  the 
square  of  ED  :  but  the  square  of  ED  is 
equals  to  the  squares  of  EB,  BD,  be- 
cause EBD  is  a  right  angle  :  therefore 
the  rectangle  AD,  DC,  together  with 
the  square  of  EB,  is  equal  to  the  squares 
of  EB,  BD  :  take  away  the  common 
square  of  EB  ;  tljerefore  the  remain- 
ing rectangle  AD,  DC  is  equal  to  the 
square  of  the  tangent  DB. 

But  if  DCA  does  not  pass  through  the  centre  of  the  circle 
ABC,  take  <*  the  centre  E,  and  draw  EF  perpendicular*  to  AC,  d  1,3. 
and  join  EB,  EC,  ED  :  and  because  the  straight  line  EF,  which  c  12. 1. 
passes  through  the  centre,  cuts  the  straight  lirie  AC,  which  does 
not  pass  through  the  centre,  at   right  ~ 

angles,     it    shall   likewise   bisect  *^  it ;  Jfi  f3. 3. 

therefore  AF  is  equal  to  PC  s  and  be- 
cause the  straight  line  AC  is  bisected  in 
F,  and  produced  to  D,  the  rectangle 
AD,  DC,  together  with  the  square  of 
FC,  is  equal  ^  to  the  square  of  FD  :  to 
each  of  these  equals  add  the  square  of 
FE  ;  therefore  the  rectangle  AD,  DC, 
together  with  the  squares  of  CF,  FE,  is 
equal  to  the  squares  of  DF,  FE  ;  but  the 
square  of  ED  is  equal  «=  to  the  squares 
of  DF,  FE,  because  Et'D  is  a  right  an- 
gle ;  and  the  square  of  EC  is  equal  to 

the  squares  of  CF,  FE  ;  therefore  the  rectangle  AD,  DC»  to- 
gether with  the  square  of  EC,  is  equal  to  the  square  of  ED : 
and  CE  is  equal  to  EB  ;  therefore  the  rectangle  AD,  DC,  tO' 
gether  with  the  square  of  EB,  is  equal  to  the  square  of  ED : 
but  the  squares  of  EB,  BD  are  equal  to  the  square  «  of  ED,  be- 


100 


THE  ELEMENTS 


Book  III.  cause  EBD  is  a  right  angle ;  therefore  the  rectangle  AD,  DC, 
U-.^— ^  together  with  the  square  of  EB,  is  equal  to  the  squares  of  EB, 
BD  :  take  away  the  common  square  of  EB  ;  therefore  the  re- 
maining rectangle  AD,  DC  is  equal  to   the   square   of  DB. 
Wherefore,  if  from  any  point,  &c.     Q.  E.  D. 

Cor.  If  from  any  point  without  a  A 

circle,  there  be  drawn  two  straight 
lines  cutting  it,  as  AB,  AC,  the  rect- 
angles contained  by  the  whole  lines 
and  the  parts  of  them  without  the 
circle,  are  equal  to  one  another,  viz. 
the  rectangle  B  A,  AE  to  the  rectan- 
gle CA,  AF  :  for  each  of  them  is 
equal  to  the  square  of  the  straight 
line  AD  which  touches  the  circle. 


PROP.  XXXVII.    THEOR. 


See  U.  IF  from  a  point  without  a  circle  there  be  drawn 
two  straight  lines,  one  of  which  cuts  the  circle,  and 
the  other  meets  it ;  if  the  rectangle  contained  by  the 
whole  line  which  cuts  the  circle,  and  the  part  of  it 
without  the  circle  be  equal  to  the  square  of  the  line 
which  meets  it,  the  Jine  which  meets  shall  touch  the 
circle. 


a  17.  3. 
b  18.  3. 

c  36.  3. 


Let  any  point  D  be  taken  without  the  circle  ABC,  and  from  it 
let  two  straight  lines  DCA  and  DB  be  drawn,  of  which  DCA 
cuts  the  circle,  and  DB  meets  it ;  if  the  rectangle  AD,  DC  be 
equal  to  the  square  of  DB,  DB  touches  the  circle. 

Draw  a  the  straight  line  DE  touching  the  circle  ABC,  find 
its  centre  F,  and  join  FE,  FB,  FD  ;  then  FED  is  a  right  •>  an- 
gle :  and  because  DE  touches  the  circle  ABC,  and  DCA  cuts 
it,  the  rectangle  AD,  DC  is  equal  ^  to  the  square  of  DE :  but 
the  rectangle  AD,  DC  is,  by  hypothesis,  equal  to  the  square  of 
DB  :  therefore  the  square  of  DE  is  equal  to  the  square  of  DB  ; 
And  the  straight  line  DE  equal  to  the  straight  line  DB  ;  and 


OF  EUCLID. 


101 


FE  is  equal  to  FB,  wherefore  DE,  EF  are  equal  to  DB,  BFj  Book  III. 

and  the  base  FD  is  common  to  the  ~ 

two  triangles  DEF,  DBF;    therefore 

the  angle  DEF  is  equal  ^  to  the  angle  / 1     \  d  8. 1. 

DBF;    but   DEF   is  a  right   angle, 

therefore  also  DBF  is  a  right  angle : 

and  FB,  if  produced,  is  a  diameter, 

and  the  straight  line  which  is  drawn 

at  right  angles  to  a  diameter,  from  the 

extremity  of  it,  touches  ^  the  circle :         i         /     ^i^  \      g  16.  3. 

therefore  DB  touches  the  circle  ABC. 

Wherefore,    if    from    a    point,    &c. 

Q.  E.  D. 


fHE 


ELEMENTS  OF  EUCLID, 


BOOK  IV. 


DEFINITIONS. 


SeeN. 


I. 

Book  IV.  ^  RECTILINEAL  figure  is  said  to  be  inscribed  in  another 
rectilineal  figure,  when  all  the  angles  of  the  inscribed  figure 
are  upon  the  sides  of  the  figure  in  which  it  is 
inscribed,  each  upon  each. 

n. 

In  like  manner,  a  figure  is  said  to  be  described 
about  another  figure,  when  all  the  sides  of 
the  circumscribed  figure  pass  through  the  an- 
gular points  of  the  figure  about  which  it  is  described,  each 
through  each* 

III. 

A  rectilineal  figure  is  said  to  be  inscribed 
in  a  circle,  when  all  the  angles  of  the  in- 
cribed  figure  are  upon  the  circumference 
of  the  circle. 

IV. 

A  rectilineal  figure  is  said  to  be  described  about  a  circle,  when 
each    side    of   the    circumscribed    figure 
touches  the  circumference  of  the  circle. 
V. 

In  like  manner,  a  circle  is  said  to  be  inscrib- 
ed in  a  rectilineal  figure,  when  the  cir- 
cumference of  the  circle  touches  each  side 
of  the  figure. 


OF  EUCLID. 


103 


VI. 

A  circle  is  said  to  be  described  about  a  rec- 
tilineal figure,  when  the  circumference  of 
the  circle  passes  through  all  the  angular 
points  of  the  figure  about  which  it  is  de- 
scribed. 

VII. 

A  straight  line  is  said  to  be  placed  in  a  circle,  when  the  extremi- 
ties of  it  are  in  the  circumference  of  the  circle. 


fiooklV. 


PROP.  I.    PROB. 

IN  a  given  circle  to  place  a  straight  line,  equal  to 
a  given  straight  line  not  greater  than  the  diameter  of 
the  circle. 


Let  ABC  be  the  given  circle,  and  D  the  given  straight  line, 
not  greater  than  the  diameter  of  the  circle. 

Draw  BC  the  diameter  of  the  circle  ABC  ;  then,  if  BC  is 
equal  to  D,  the  thing  required  is  done  ;  for  in  the  circle  ABC 
a  straight  line  BC  is  placed 
equal  to  D  ;  but,  if  it  is  not,  BC 
is  greater  than  D  ;  make  CE 
equal » to  D,  and  from  the  cen- 
tre C,  at  the  distance  CE,  de- 
scribe the  circle  AEF,  and  join 
CA ;  therefore,  because  C  is 
the  centre  of  the  circle  AEF, 
CA  is  equal  to  CE  ;  but  D  is 
equal  to  CE;  therefore  D  is 
equal  to  CA  :    wherefore,    in 

the  circle  ABC,  a  straight  line  is  placed  equal  to  the  given  straight 
line  D,  which  is  not  greater  than  the  diameter  of  the  circle. 
Which  was  to  be  done. 


a  3.1. 


PROP.  II.     PROB. 


IN  a  given  circle  to  inscribe  a  triangle  equiangular 
to  a  given  triangle. 


104 


THE  ELEMENTS 


Book  IV. 


a  17.  3. 
b  23. 1. 


c  32.  3. 


Let  ABC  be  the  given  circle,  and  DEF  the  given  triangle  ;  it 
is  required  to  inscribe  in  the  circle  ABC  a  triangle  equiangular 
to  the  triangle  DEF. 

Draw  a  the  straight  line  GAH  touching  the  circle  in  the  point 
A,  and  at  the  point  A,  in  the  straight  line  AH,  make''  the  angle 
HAC  equal  to  the  angle  DEF ;  and  at  the  point  A,  in  the  straight 
line  AG,  make  the  an- 
gle GAB  equal  to  the  ^  """^""-^^  A 
angle  DFE,  and  join 
BC  :  therefore  because 
HyVG  touches  the  cir- 
cle ABC,  and  AC  is 
drawn  from  the  point 
of  contact,  the  angle 
PL\C  is  equal  c  to  the 
angle  ABC  in  the  alter- 
nate segment  of  the  cir- 
cle :  but  HAC  is  equal  to  the  angle  DEF  ;  therefore  also  the  an- 
gle ABC  is  equal  to  DEF  :  for  the  same  reason,  the  angle  ACB 
is  equal  to  the  angle  DFE  ;  therefore  the  remaining  angle  BAC 
is  equal ^  to  the  remaining  angle  EDF :  wherefore  the  triangle 
ABC  is  equiangular  to  the  triangle  DEF,  and  it  is  inscribed  in 
the  circle  ABC.     Which  was  to  be  done. 


PROP.  HI.     PROB. 


ABOUT  a  given  circle  to  describe  a  triangle  equi- 
angular to  a  given  triangle. 


Let  ABC  be  the  given  circle,  and  DEF  the  given  triangle ;  it 
is  required  to  describe  a  triangle  about  the  circle  ABC  equiangu- 
lar to  the  triangle  DEF. 

Produce  EF  both  ways  to  the  points  G,  H,  and  find  the  centre 
K  of  the  circle  ABC,  and  from  it  draAV  any  straight  line  KB ; 

<i  23.  1.  at  the  point  K,  in  the  straight  line  KB,  make^  the  angle 
BKA  equal  to  the  angle  DEG,  and  the  angle  BKC  equal  to  the 
angle  DFH  ;  and  through  the  points  A,  B,  C  draw  the  straight 

b  17. 3.  lines  LAM,  MBN,  NCL  touching  ^  the  circle  ABC  :  there- 
fore because  LM,  MN,  NL  touch  the  circle  ABC  in  the 
points  A,  B,  C,  to  which  from  the  centre  are  drawn  KA,  KB, 

c  18.3.  KC,  the  angles  at  the  points  A,  B,  C  are  rights  angles:  and 
because  the  four  angles  of  the  quadrilateral  figure  AMBK  are 


OF  EUCLID. 


105 


d  13. 1. 


equal  to  four  right  angles,  for  it  can  be  divided  into  two  tri-BooklV. 
angles :  and  that  two  of  them  KAM,  KBM  are  right  angles,  the 
other  two  AKB, 
AMB  are  equal  to 
two  right  angles  : 
but  the  angles 
DEG,  DEF  are 
likewise  equaH  to 
two  right  angles ; 
therefore  the  an- 
gles AKB,  AMB 
are  equal  to  the 
angles  DEG,  DEF 
of  which   AKB  is     ^^  B  N 

equal  to  DEG ;  wherefore  the  remaining  angle  AMB  is  equal 
to  the  remaining  angle  DEF :  in  like  manner,  the  angle  LNM 
may  be  demonstrated  to  be  equal  to  DFE ;  and  therefore  the 
remaining  angle  MLN  is  equal  e  to  the  remaining  angle  EDF :  e  32. 1. 
wherefore  the  triangle  LMN  is  equiangular  to  the  triangle 
DEF :  and  it  is  described  about  the  circle  ABC.  Which  was  to 
be  done. 


PROP.  IV.    PROB. 


TO  inscribe  a  circle  in  a  given  triangle. 


SeeN. 


Let  the  given  triangle  be  ABC ;  it  is  required  to  inscribe  a 
circle  in  ABC. 

Bisect  a  the  angles  ABC,  BC  A  by  the  straight  lines  BD,  CD  »  9- 1- 
meeting  one  another  in  the  point  D,  from  which  draw ''  DE, ''  ^2- 1. 
DF,  DG  perpendiculars  to  AB,  A 

BC,  CA :  and  because  the  angle 
EBD  is  equal  to  the  angle  FBD, 
for  the  angle  ABC  is  bisected  by 

BD,  and  that  the  right  angle 
BED  is  equal  to  the  right  angle 
BFD,  the  two  triangles  EBD, 
FBD  have  two  angles  of  the  one 
equal  to  two  angles  of  the  other, 
and  the  side  BD,  which  is  oppo- 
site to  one  of  the  equal  angles  in 
each,  is  common  to  both  ;  there- 
fore their   other  sides  shall  be 

O 


106 


THE  ELEMENTS 


Book  IV.  equal  c ;  wherefore  DE  is  equal  to  DF :  for  the  same  reason, 
v.^vr"-*>  DG  is  equal  to  DF ;  therefore  the  three  straight  lines  DE,  DF, 
c  26. 1.     DG  are  equal  to  one  another,  and  the  circle  described  from  the 
centre  D,  at  the  distance  of  any  of  them,  shall  pass  through  the 
extremities  of  the  other  two,   and  touch  the  straight  lines  AB, 
BC,  CA,  because  the  angles  at  the  points  E,  F,  G  are  right  an- 
gles, and  the  straight  line  which  is  drawn  from  the  extremity  of 
d  16.  3.     a  diameter  at  right  angles  to  it,  touches 'l  the  circle:  therefore 
the  straight  lines  AB,  BC,  CA  do  each  of  them  touch  the  circle, 
and  the  circle  EFG  is  inscribed  in  the  triangle  ABC.     Which 
was  to  be'done. 


PROP.  V.    PROB. 


See  N.        TO  describe  a  circle  about  a  given  triangle. 


Let  the  given   triangle  be  ABC ;  it  is  required  to  describe  a 
circle  about  ABC. 
a  10. 1.         Bisect  a  AB,  AC  in  the  points  D,  E,  and  from  these  points 
b  11. 1.     draw  DF,  EF  at  right  angles'*  to  AB,  AC;  DF,  EF  produced 


V> 


C  B 


c4. 1. 


meet  one  another:  for,  if  they  do  not  meet,  they  are  parallel, 
wherefore  AB,  AC,  Avhich  are  at  right  angles  to  them,  are  pa- 
rallel ;  which  is  absurd :  let  them  meet  in  F,  and  join  FA  ; 
also,  if  the  point  F  be  not  in  BC,  join  BF,  CF :  then,  because 
AD  is  equal  to  DB,  and  DF  common,  and  at  right  angles  to 
AB,  the  base  AF  is  ecjual^  to  the  base  FB:  in  like  manner,  it 
may  be  shown,  that  CF  is  equal  lo  FA ;  and  therefore  BF  is 
equal  to  IC;    and   FA,   FB,  FC   are  equal  to  one  anothei: ; 


OF  EUCLID. 


lor 


wherefore  the  circle  described  from  the  centre  F,  at  the  distance  Book  IV. 
of  one  of  them,  shall  pass  through  the  extremities  of  the  other  *'■— r— ^ 
two,  and  be  described  about  the  triangle  ABC.  Which  was  to 
be  done. 

Cor.  And  it  is  manifest,  that  when  the  centre  of  the  circle 
falls  within  the  triangle,  each  of  its  angles  is  less  than  a  right 
angle,  each  of  them  being  in  a  segment  greater  than  a  semicir- 
cle ;  but,  when  the  centre  is  in  one  of  the  sides  of  the  triangle, 
the  angle  opposite  to  this  side,  being  in  a  semicircle,  is  a  right 
angle  ;  and,  if  the  centre  falls  without  the  triangle,  the  angle  op- 
posite to  the  side  beyond  which  it  is,  being  in  a  segment  less 
than  a  semicircle,  is  greater  than  a  right  angle  :  wherefore,  if  the 
given  triangle  be  acute  angled,  the  centre  of  the  circle  falls  within 
tt;  if  it  be  a  right  angled  triangle,  the  centre  is  in  the  side  oppo- 
site to  the  right  angle  ;  and,  if  it  be  an  obtuse  angled  triangle,  the 
centre  falls  without  the  triangle,  beyond  the  side  opposite  to  the 
obtuse  angle. 


PROP.  VI.    PROB. 


TO  inscribe  a  square  in  a  given  circle. 


ft  4.1. 


Let  ABCD  be  the  given  circle  ;  it  is  required  to  inscribe  a 
square  in  ABCD. 

Draw  the  diameters  AC,  BD  at  right  angles  to  one  another ; 
and  join  AB,  BC,  CD,  DA  ;  because  BE  is  equal  to  ED,  for  E  is 
the  centre,  and  that  EA  is  common,  A 

and  at  right  angles  to  BD  ;  the  base 
BA  is  equal  ^  to  the  base  AD  ;  and,  for 
the  same  reason,  BC,  CD  are  each  of 
them  equal  to  BA  or  AD  ;  therefore 
the  quadrilateral  figure  ABCD  is  equi- 
lateral.    It  is  also  rectangular  ;  for  the 
straight  line  BD,  being  the  diameter 
of  the  circle  ABCD,  BAD  is  a  semi- 
circle ;  wherefore  the  angle  BAD  is  a 
right  ^  angle ;  for  the  same  reason  each  of  the  angles  ABC,  BCD,  b  31. 3. 
CDA  is  a  right  angle  ;  thei*efore  the  quadrilateral  figure  ABCD 
is  rectangular,  and  it  has  been  shown  to  be  equilateral ;  therefore 
it  is  a  square  ;  and  it  is  inscribed  in  the  circle  ABCD.     Which 
was  to  be  done. 


Book  IV. 


THE  ELEMENTS 


a  17.  3. 


b  18. 3. 


c  28. 1. 


d34.1. 


PROP.  VII.    PROB. 
TO  describe  a  square  about  a  given  circle. 

Let  ABCD  be  the  given  circle ;  it  is  required  to  describe  a 
square  about  it. 

Draw  two  diameters  AC,  BD  of  the  circle  ABCD,  at  right  an- 
gles to  one  another,  and  through  the  points  A,  B,  C,  D  draw  * 
FG,  GH,  HK,  KF  touching  the  circle ;  and  because  FG  touches 
the  circle  ABCD,  and  EA  is  drawn  from  the  centre  E  to  th& 
point  of  contact  A,  the  angles  at  A  are  right  ^  angles  ;  for  the 
same  reason,  the  angles  at  the  points  B,  C,  D  are  right  angles ; 
and  because  the  angle  AEB  is  a  right       G  A  F 

angle,  as  likewise  is  EBG,  GH  is  pa- 
rallel "^  to  AC  ;  for  the  same  reason, 
AC  is  parallel  to  FK,  and  in  like  man- 
ner GF,  HK  may  each  of  them  be  de- 
monstrated to  be  parallel  to  BED  ; 
therefore  the  figures  GK,  GC,  AK, 
FB,  BK  are  parallelograms  ;  and  GF 
is  therefore  equal  ^  to  HK,  and  GH  to 

FK  ;  and  because  AC  is  equal  to  BD,       H  C  K 

and  that  AC  is  equal  to  each  of  the  two  GH,  FK  ;  and  BD  to 
each  of  the  two  GF,  HK :  GH,  FK  are  each  of  them  equal  to 
GF  or  HK  ;  therefore  the  quadrilateral  figure  FGHK  is  equilate- 
ral. It  is  also  rectangular  ;  for  GBEA  being  a  parallelogram, 
and  AEB  a  right  angle,  AGB  ^  is  likewise  a  right  angle  :  in  the 
same  manner,  it  may  be  shown  that  the  angles  at  H,  K,  F  are 
right  angles  ;  therefore  the  quadrilateral  figure  FGHK  is  rect- 
angular, and  it  was  demonstrated  to  be  equilateral  ;  therefore  it 
is  a  square  ;  and  it  is  described  about  the  circle  ABCD.  Which 
was  to  be  done. 


B 


r. 

N 

J 

D 


PROP.  VIII.    PROB. 

TO  inscribe  a  circle  in  a  given  square. 

I^et  ABCD  be  the  given  square  ;  it  is  required  to  inscribe  a 
circle  in  ABCD. 
a.  10. 1.        Bisect  ^  each  of  the  sides  AB,  AD,  in  the  points  F,  E,  and 
b31  1.    through  E  draw  •»  EH  parallel  to  AB  or  DC,  and  through  F 


OF  EUCLID. 


109 


draw  FK  parallel  to  AD  or  BC ;   therefore  each  of  the  figures  Book  IV. 
AK,  KB,  AH,  HD,  AG,  GC,  BG,  GD  is  a  parallelogram,  and  their  ^— r— ' 
opposite  sides  are  equal^  ;  and  because  AD  is  equal  to  AB,  and  c  34. 1. 
that  AE  is  the  half  of  AD,  and  AF  the  half  of  AB,  AE  is  equal 
to    AF;    wherefore   the    sides   opposite     A  E  D 

to  these  are  equal,  viz.  FG  to  GE ;  in 
the  same  manner,  it  may  be  demon- 
strated that  GH,  GK  are  each  of  them 
equal  to  FG  or  GE ;  therefore  the 
four  straight  lines  GE,  GF,  GH,  GK 
are  equal  to  one  another ;  and  the  cir- 
cle described  from  the  centre  G,  at  the 
distance  of  one  of  them,  shall  pass 
through  the  extremities  of  the  other 
three,  and  touch  the  straight  lines  AB, 

BC,  CD,  DA :  because  the  angles  at  the  points  E,  F,  H,  K  are 
right  "i  angles;  and  that  the  straight  line  which  is  drawn  from  d 
the  extremity  of  a  diameter,  at  right  angles  to  it,  touches  the 
circle  «  ;  therefore  each  of  the  straight  lines  AB,  BC,  CD,  DAe  16.  3. 
touches  the  circle,  which  therefore  is  inscribed  in  the  square 
ABCD.     Which  was  to  be  done. 


iJ 


B 


H 


C 


29.1. 


PROP.  IX.    PROB. 


TO  describe  a  circle  about  a  given  square. 


a  8.1. 


Let  ABCD  be  the  given  square;  it  is  required  to  describe  a 
circle  about  it. 

Join  AC,  BD  cutting  one  another  in  E  ;  and  because  DA  is 
equal  to  AB,  and  AC  common  to  the  triangles  DAC,  BAC, 
the  two  sides  DA,  AC  are  equal  to  the 
two  BA,  AC ;  and  the  base  DC  is  equal  ^ 
to  the  base  BC  ;  wherefore  the  angle 
DAC  is  equal*  to  the  angle  BAC,  andi 
the  angle  DAB  is  bisected  by  the  straight] 
line  AC :  in  the  same  manner,  it  may  be 
demonstrated  that  the  angles  ABC,  BCD, 
CDA  are  severally  bisected  by  the  straight  ^^ 
lines  BD,  AC ;  therefore,  because  the 
angle  DAB  is  equal  to  the  angle  ABC,  and  that  the  angle 
EAB  is  the  half  of  DAB,  and  EBA  the  half  of  ABC;  the 
angle  EAB  is  equal  to  the  angle  EBA;  wherefore  the  side 
EA  is  equal  *»  to  the  side  EB  :   in  the  same  manner,  it  may  be  b  6.1. 


HO 


TITE  ELEMENTS 


Book  IV.  demonstrated,  that  the  strais^ht  lines  EC,  ED  are  each  of  them 
*— "v^-*  equal  to  EA  or  EB ;  therefore  the  four  straight  lines  EA,  EB, 
EC,  ED  are  equal  to  one  another ;  and  the  circle  described  from 
the  centre  E,  at  the  distance  of  one  of  them,  shall  pass  through 
the  extremities  of  the  other  three,  and  be  described  about  the 
square  ABCD.     Which  was  to  be  done. 


PROP.  X.    PROD. 

TO  describe  an  isosceles  triangle,  having  each  of 
the  angles  at  the  base  double  of  the  third  angle. 


all.  2.         Take  any  straight  line  AB,  and  divide^  it  in  the  point  C,  so 

that  the  rectangle  AB,  BC  be  equal  to  the  square  of  CA ;  and 

from  the  centre  A,  at  the  distance  AB,  describe  the  circle  BDE, 
b  1.  4.      in  which  place  ^  the  straight  line  BD  equal  to  AC,  which  is  not 

greater  than  the  diameter  of  the  circle  BDE ;  join  DA,  DC,  and 
c  5.  4.       about  the  triangle  ADC  describe  =  the  circle  ACD  ;  the  triangle 

ABD  is   such  as  is  required,  that  is,  each  of  the  angles  ABD, 

ADB  is  double  of  the  angle  BAD. 

Because  the  rectangle  AB,  BC  is  equal  to  the  square  of  AC, 

and  that  AC  is  equal  to  BD,  the  rectangle  AB,  BC  is  equal  to 

the  square  of  BD  ;  and  because 

from  the  point  B,    without  the 

circle  ACD,   two  straight  lines 

BCx\,  BD  are  drawn  to  the  cir- 
cumference, one  of  which  cuts, 

and  the  other  meets  the  circle, 

and  that  the  rectangle  AB,  BC 

contained   by  tlie  whole  of  the 

cutting  line,  and  the  part  of  it 

without  the  circle,  is  equal  to  the 

square  of  BD  which  meets  it; 
d  ^r.  o.    the   straight  line  BD  touches 'i 

the  circle   ACD  ;   and  because 

BD  touches  the  circle,  and  DC 

is  drawn  from  the  point  of  con- 
e  33.  3.    tact  D,  the   angle  BDC  is  equals  to  the  angle  DAC  in  the 

alternate  segment  of  the  circle;  to  each  of  these  add  the  angle 

CDA;    therefore    the  whole  angle  BDA  is    equal    to   the   two 
f  .32.  1.     angles  CDA,  DAC;  but  the   exterior   angle  BCD  is  equal  f  to 

the  angles  CD.\,  DAC ;  therefore  also  BDA  is  equ&l  to  BCD  ; 


OF  EUCLID.  Ill 

but  BDA  is  equals  to  the  aiif^le  CBD,  because  the  side  AD  Book IV. 
is  equal  to  the  side  AB  ;  therefore  CBD,  or  DBA  is  equal  to  *»— v— ' 
BCD  ;    and  consequently  thT  three  angles  BDA,  DBA,  BCD  g  5. 1. 
are  equal  to  one  another  ;  and  because  the  angle  DBC  is  equal 
to  the  angle  BCD,  the  side  BD  is  equal  ^  to  the  side  DC  ;  but  h  6. 1 
BD  was  made  equal  to  CA ;  therefore  also  CA  is  equal  to  CD, 
and  the  angle  CDA  equals  to  the  angle  DAC  ;  therefore  the 
angles  CD  A,  D  AC   together,  are    double    of  the    angle   DAC: 
but  BCD  is  equal  to  the  angles  CDA,  DAC  ;  therefore  alsc^  BCD 
is  double  of  DAC,  and  BCD  is  equal  to  each  of  the  angles  BDA, 
DBA  ;  each  therefore  of  the  angles  BDA,  DBA  is  double  of  the 
angle  DAB;  wherefore  an  isosceles  triangle  ABD  is  described, 
having  each  of  the  angles  at  the  base  double  of  the  third  angle. 
Which  was  to  be  done. 


PROP.  XI.    PROB. 


TO  inscribe  an  equilateral  and  equiangular  penta- 
gon in  a  given  circle. 

Let  ABCDE  be  the  given  circle  ;  it  is  required  to  inscribe  an 
equilateral  and  equiangular  pentagon  in  the  circle  ABCDE. 

Describe*  an   isosceles  triangle  FGH,    having  each    of   the  a  10. 4 
angles  at  G,  H,  double  of  the   angle  at   F  ;   and  in  the  circle 
ABCDE  inscribe  ^  the  triangle  ACD   equiangular  to   the  tri-  b  2.  4 
angle  FGH,  so  that  the  angle  A 

CAD  be  equal  to  the  angle 
at  F,  and  each  of  the  angles 
ACD,  CDA  equal  to  the 
angle  at  G  or  H  ;  wherefore 
each  of  the  angles  ACD, 
CDA  is  double   of  the  angle 

CAD.      Bisect  c    the     angles      /       \      \\  /  ^<C^  \  //       c  9. 1 
ACD,  CDA  by    the    straight 
lines  CE,  DB  ;  and  join  AB, 
BC,    DE,    EA.     ABCDE    is 
the  pentagon  required. 

Because  each  of  the    angles   ACD,  CDA  is  double  of  CAD, 
and   are  bisected  by  the  straight  lines  CE,  DB,  the   five  angles 
DAC,   ACE,  ECD,  CDB,  BDA  are   equal  to  one  another ;  but 
equal  angles  stand  upon    equal  <i  circumferences;  therefore  (he  d  26.  5. 
five  circumferences  AB,  BC,  CD,  DE,  EA  are   eoual  to  one 


il2  THE  ELEMENTS 

Book  IV.  another :   and  ec'ual  circumferences  are  subtended   by  equal « 
^-—v^^  straight  lines  ;    therefore  the  five   straight  lines  AB,    BC,  CD, 
c  29.  3.     DE,  EA  are  equal  to  one   another.     Wherefore   the  pentagon 
ABCDE   is  equilateral.     It  is   also  equiangular ;    because  the 
circumference  AB  is  equal  to  the  circumference  DE  :  if  to  each 
be  added  BCD,  the  whole  ABCD  is  equal  to  the  whole  EDCB  : 
and   the  angle  AED   stands  on  the  circumference  ABCD,  and 
the    angle  BAE  on  the  circumference    EDCB;    therefore   the 
£27.3.     angle  BAE    is  equal  f  to  the  angle  AED:  for  the  same   reason, 
each  of  the  angles  ABC,  BCD,  CDE  is  equal  to  the  angle  BAE, 
or  AED  :  therefore  the  pentagon  ABCDE  is  equiangular ;    and 
it  has  been  shown  that  it  is  equilateral.     Wherefore,  in  the  given 
circle,  an  equilateral  and  equiangular  pentagon  has  been  inscrib- 
ed.    ^Viuch  was  to  be  done. 


PROP.  XII.     PROB. 


TO  describe  an  equilateral  and  equiangular  penta- 
gon about  a  given  circle. 

Let  ABCDE  be  the  given  circle  ;  it  is  required  to  describe 

an  equilateral  and  equiangular  pentagon  about  the  circle  ABCDE. 

Let  the  angles  of  a  pentagon,  inscribed  in  the  circle,  by   the 

last  proposition,  be  in  the  points  A,  B,  C,  D,  E,  so  that  th^ 

a  11  4.  circumferences  AB,  BC,  CD,  DE,  EA  are  equal  »  ;  and  through 
the   points  A,    B,  C,   D,   E  draw  GH,   HK,  KL,   LM,  MG, 

b  17.  3.  touching •>  the  circle;  take  the  centre  F,  and  join  FB,  FK,  FC, 
FL,  FD :  and  because  the  straight  line  KL  touches  the  circle 
ABCDE  in  the   point   C,    to   which    FC   is    drawn  from  the 

c  18.  3.  centre  F,  FC  is  perpendicular  =  to  KL  ;  therefore  each  of  the 
angles  at  C  is  a  right  angle  :  for  the  same  reason,  the  angles  at 
the  points    B,  D,    are    right  angles :    and    because  FCK    is  a 

d  47.  1.  right  angle,  the  square  of  FK  is  equal  ^  to  the  squares  of  FC, 
CK  :  for  the  same  reason,  the  square  of  FK  is  equal  to  the  squares 
of  FB,  BK  :  therefore  the  squares  of  FC,  CK  are  equal  to  the 
squares  of  FB,  BK,  of  which  the  square  of  FC  is  equal  to  the 
squa:-:;  of  FB ;  tl.c  remaining  square  of  CK  is  therefore  equal  to 


OF   EUCLID. 


113 


the  remaining  square  of  BK,  and  the  straight  line  CK  equal  to  Book  IV. 
BK :  and  because  FB  is  equal  to  FC,  and  FK  common  to  the  ^— -y— ^ 
triangles  BFK,  CFK,  the  two  BF,  FK  are  equal  to  the  two  CF, 
FK ;  and  the  base  BK  is  equal  to  the  base  KC ;  therefore  the 
angle  BFK  is  equal  e  to  the  angle  KFC,  and  the  angle  BKF  to  e  8.  1. 
FKC ;  wherefore  the  angle  BFC  is  double  of  the  angle   KFC, 
and  BKC  double  of  FKC  ;  for  the  same  reason,  the  angle  CFD 
is  doublp  of  the  angle  CFL,  and  CLD  double  of  CLF  :  and  be- 
cause the  circumference  BC  is  equal  to  the  circumference  CD, 
the  angle  BFC  is  equal  f  to  the  •  f  27.  3. 

angle  CFD;  and  BFC  is  dou-  G 

ble  of  the  angle  KFC,  and  CFD 
double  of  CFL ;  therefore  the 
angle  KFC  is  equal  to  the  angle 
CFL  ;  and  the  right  angle  FCK 
is  equal  to  the  right  angle  FCL: 
therefore,  in  the  two  triangles 
FKC,  FLC,  there  are  two  an- 
gles of  one  equal  to  two  angles 
of  the  other,  each  to  each,  and 
the  side  FC,  which  is  adjacent 
to  the  equal  angles  in  each,  is 
common  to  both  ;  therefore  the 
other  sides  shall  be  equal  s  to  the  other  sides,  and  the  third  angle  S  26. 1 
to  the  third  angle  :  therefore  the  straight  line  KC  is  equal  to  CL, 
and  the  angle  FKC  to  the  angle  FLC  :  and  because  KC  is  equal  to 
CL,  KL  is  double  of  KC  :  in  the  same  manner,  it  may  be  shown  that 
HK  is  double  of  EK :  and  because  BK  is  equal  to  KC,  as  was 
demonstrated,  and  that  KL  is  double  of  KC,  and  HK  double  of 
BK,  HK  shall  be  equal  to  KL :  in  like  manner  it  may  be  shown 
that  GH,  GM,  ML  are  each  of  them  equal  to  HK  or  KL  :  there- 
fore the  pentagon  GHKLM  is  equilateral.  It  is  also  equiangular ; 
for,  since  the  angle  FKC  is  equal  to  the  angle  FLC,  and  that  the 
angle  HKL  is  double  of  the  angle  FKC,  and  KLM  double  of  FLC, 
as  was  before  demonstrated,  the  angle  HKL  is  equal  to  KLM: 
and  in  like  manner  it  may  be  shown,  that  each  of  the  angles  KHG, 
HGM,  GML  is  equal  to  the  angle  HKL  or  KLM :  therefore  the 
five  angles  GHK,  HKL,  KLM,  LMG,  MGH  being  equal  to  one 
another,  the  pentagon  GHKLM  is  equiangular :  and  it  is  equila- 
teral, as  was  demonstrated ;  and  it  is  described  about  the  circle 
ABCDE.     Which  Avas  to  be  done. 


114 
Book  IV. 


THE  ELEMENTS 


PROP.  XIII.    PROB. 


A 


TO  inscribe  a  circle  in  a  given  equilateral  and  equi- 
angular pentagon. 

Let  ABCDE  be  the  given  equilateral  and  equiangular  penta- 
gon ;  it  is  required  to  inscribe  a  circle  in  the  pentagon  ABCDE. 
a  9.1.  Bisect  2  -.he  angles  BCD,  CDE  by  the  straight  lines  CF,  DF, 

and  from  the  point  F,  in  which  they  meet,  draw  the  straight  lines 
FB,  FA,  FE :  therefore,  since  BC  is  equal  to  CD,  and  CF  com- 
mon to  the  triangles  BCF,  DCF,  the  two  sides  BC,  CF  are  equal 
to   the  two  DC,  CF ;  and  the  angle  BCF  is  equal  to  the  angle 
b  4.  1.      DCF  ;  therefore  the  base  BF  is  equal  ^  to  the  base  FD,  and  the 
other  angles  to  the  other  angles,  to  which  the  ecjual  sides  are  op- 
posite ;  therefore  the  angle  CBF  is  equal  to  the  angle  CDF  :  and 
because  tiie  angle  CDE  is  double  of  CDF,  and  that  CDE  is  equal 
to  CBA,  and  CDF  to  CBF  ;  CBA 
is  also  double  of  the  angle  CBF  ; 
therefore  the  angle  ABF  is  equal 
to  the  angle  CBF;   wherefore  the 
angle    ABC    is    bisected    by    the 
straight   line    BF :    in   the    same 
manner,  it  may  be  demonstrated, 
that  the  angles  BAE,   AED  are 
bisected  by  the  slraigiit  lines  AF, 
c  12. 1.     FE  :    froui    the   point   F    draw  c 
FG,  FH,  FK,    FL,    FM  perpen- 
diculars to  the  straight  lines  AB, 
BC,    CD,     DE,     EA:    and    be- 
cause tile  angle  IICF  is  equal  to 
KCF,  and  tiie  right  angle  FHC  equal  to  the  right  angle  FKC  ; 
in  liie  triangles  FHC,  FKC  there  are  two  angles  of  one  equal 
to  two  anj^;les  of  the  other,  and  the  side  FC,   which  is  opposite 
to  one  of  the  equal  angles  in  each,  is  common  to  both  ;  therefore 
d  26.  1.    tlie  otiier  sides  shall  be  eciual 'i,   each  to  each;    wherefore    the 
perpendicular  FII  is  equal  to  tiie  perpendicular  FK :  in  the  same 
manner  it  may  be  demonstrated  that  FL,  FM,  FG  are  each  pf 
them  equal  to  FII  or  FK ;  therefore  the  five  straight  lines  FG, 
I'll,  FK,  FL,  FM  are  equal  to  one  another  :  wherefore  the  cir- 
cle described  from  the  centre  F,   at  the  distance  of  one  of  these 
live,  shall  pass  through  the  extremities  of  the  other  four,  and 


OF  EUCLID. 


115 


touch  the  straight  lines  AB,  BC,  CD,  DE,  EA,  because  the  Book  IV. 
angles  at  the  points  G,  H,  K,  L,  M  are  right  angles  ;  and  that  ^— v— ' 
a  straight  line  drawn  from  the  extremity  of  the  diameter  of  a 
circle  at  right  angles  to  it,  touches  e  the  circle  :  therefore  each  e  15.  3- 
of  the  straight  lines  AB,  BC,  CD,  DE,  EA  touches  the  circle  ; 
wherefore  it  is  inscribed  in  the  pentagon  ABCDE.     Which  was 
to  be  done. 


PROP.  XIV.    PROB. 


TO  describe  a  circle  about  a  given  equilateral  and 
equiangular  pentagon. 

Let  ABCDE  be  the  given  equilateral  and  equiangular  penta- 
gon ;  it  is  required  to  describe  a  circle  about  it. 

Bisect  a  the  angles  BCD,  CDE  by  the  straight  lines  CF,  FD,  a  9. 1. 
and  from  the  point  F,  in  which  they  meet,  draw  the  straight 
lines  FB,  FA,  FE  to  the  points  B, 
Aj  E.  It  may  be  demonstrated,  in 
the  same  manner  as  in  the  preceding 
proposition,  that  the  angles  CBA, 
BAE,  AED  are  bisected  by  the 
straight  lines  FB,  FA,  FE :  and 
because  the  angle  BCD  is  equal  to 
the  angle  CDE,  and  that  FCD  is 
the  half  of  the  angle  BCD,  and  CDF 
the  half  of  CDE  ;  the  angle  FCD  is 
equal  to  FDC  ;  wherefore  the  side 
CF  is  equal  *>  to  the  side  FD  :  in  Hke  manner  it  may  be  demon-  b  6.  l\ 
strated  that  FB,  FA,  FE  are  each  of  them  equal  to  FC  or  FD  : 
therefore  the  five  straight  lines  FA,  FB,  FC,  FD,  FE  arc  equal 
to  one  another ;  and  the  circle  described  from  the  centre  F,  at 
the  distance  of  one  of  them,  shall  pass  through  the  extremities 
of  the  other  four,  and  be  described  about  the  equilateral  and 
equiangular  pentagon  ABCDE.     Which  wa«  to  be  done. 


THE  ELEMENTS 


PROP.  XV.    PROB. 


SeeN. 


a  5.  1. 
b  32. 1. 


C  13.  1. 


d  15.  1. 


e26. 


f  29.  .1. 


TO  inscribe  an  equilateral  and  equiangular  hexa- 
gon in  a  given  circle. 

Let  ABCDEF  be  the  given  circle  ;  it  is  required  to  inscribe 
im.  ecfuilateral  and  equiangular  hexagon  in  it. 

Find  the  centre  G  of  the  circle  ABCDEF,  and  draw  the  dia- 
meter AGD  ;  and  from  D  as  a  centre,  at  the  distance  DC,  de- 
scribe the  circle  EGCH,  join  EG,  CG,  and  produce  them  to  the 
points  B,  F  ;  and  join  AB,  BC,  CD,  DE,  EF,  FA :  the  hexagon 
ABCDEF  is  equilateral  and  equiangular. 

Because  G  is  the  centre  of  the  circle  ABCDEF,  GE  is  equal 
to  GD  :  and  because  D  is  the  centre  of  the  circle  EGCH,  DE 
is  equal  to  DG ;  wherefore  GE  is  equal  to  ED,  and  the  tri- 
angle EGD  is  equilateral;  and  therefore  its  three  angles  EGD, 
GDE,  DEG  are  equal  to  one  another,  because  the  angles  at 
the  base  of  an  isosceles  triangle  are  equal  *;  and  the  three  angles 
of  a  triangle  are  equal  b  to  two  right  angles  ;  therefore  the 
angle  EGD  is  the  third  part  of  two  right  angles:  in  the  same 
manner  it  may  be  demonstrated,  that 
the  angle  DGC  is  also  the  third  part 
of  two  right  angles :  and  because  the 
straight  line  GC  makes  with  EB  the 
adjacent  angles  EGC,  CGB  equal  <= 
to  two  right  angles  ;  the  remaining 
angle  CGB  is  the  third  part  of  two 
right  angles ;  therefore  the  angles 
EGD,  EIGC,  CGB  are  equal  to  one 
another:  and  to  these  arc  equaH  the 
vertical  opposite  angles  BGA,  AGF, 
FGE  :  therefore  the  six  angles  EGD, 
DGC,  CGB,  BGA,  AGF,  FGE  are 
equal  to  one  another :  but  equal 
angles  stand  upon  equal  «  circumfe- 
rences ;  therefore  the  six  circumfe- 
rences AB,  BC,  CD,  DE,  EF,  FA  are  equal  to  one  another : 
and  equal  circumferences  are  subtended  by  equal  *"  straight 
lines  ;  therefore  the  six  straight  lines  are  equal  to  one  another, 
and  the  hexagon  ABCDEF  is  equilateral.  It  is  also  equiangu- 
lar ;  for,  since  the  circumference  AF  is  equal  to  ED,  to  each  of 
thtfcC  add  the  circumference  ABCD  :  therefore  the  whole  cir- 
cumference FABCD  shall  be   equal   to   the    whole    EDCBA  ; 


OF  EUCLID. 


iir 


and  the  angle  FED  stands  upon  the  circumftiience  FABCD,  and  Book IV. 
the  angle  AFE  upon  EDCBA ;  therefore  the  angle  AFE  is  *— v—' 
equal  to  FED  :  in  the  same  manner  it  may  be  demonstrated  that 
the  other  angles  of  the  hexagon  ABCDEF  are  each  of  them 
equal  to  the  angle  AFE  or  FED ;  therefore  the  hexagon  is 
equiangular;  and  it  is  equilateral,  as  was  shown;  and  it  is  in- 
scribed in  the  given  circle  ABCDEF.     Which  was  to  be  done. 

Cor.  From  this  it  is  manifest,  that  the  side  of  the  hexagon  is 
equal  to  the  straight  line  from  the  centre,  that  is,  to  the  semi- 
diameter  of  the  circle. 

And  if  through  the  points  A,  B,  C,  D,  E,  F  there  be  drawn 
straight  lines  touching  the  circle,  an  equilateral  and  equiangular 
hexagon  shall  be  described  about  it,  which  may  be  demonstrat- 
ed from  what  has  been  said  of  the  pentagon  ;  and  likewise  a  cir- 
cle may  be  inscribed  in  a  given  equilateral  and  equiangular  hexa- 
gon, and  circumscribed  about  it,  by  a  method  like  to  that  used 
for  the  pentagon. 


PROP.  XVI.     PIIOB. 

TO  inscribe  an  equilateral  and  equiangular  quin-  See  n. 
decagon  in  a  given  circle. 


Let  ABCD  be  the  given  circle  ;    it  is  required  to  inscribe  an 
equilateral  and  equiangular  quindecagon  in  the  circle  ABCD. 

Let  AC  be  the  side  of  an  equilateral  triangle  inscribed  ^  in  a  2.  4. 
the  circle,  and  AB  the  side  of  an  equilateral   and  equiangular 
pentagon  inscribed  ^  in  the  same  ;   therefore,  of  such  equal  parts  b  11.  4- 
as  the  whole  circumference   ABCDF   contains  fifteen,  the  cir- 
cumference ABC,  being  the  third  \ 
part  of  the  whole,  contains  five  ;  and 
the  circumference  AB,  which  is  the 
fifth  part  of   the   whole,   contains 
three  ;  therefore  BC  their  difference 
contains  two  of  the  same  parts:  bi- 
sect ^BC  in  E;   therefore  BE,  EC 
are,  each  of  them,  the  fifteenth  part 
of  the  whole  circumference  ABCD  : 
tlicrefore,  if  the  straight  lines  BE, 
EC   be   drawn,   and   straight   lines 
equal  to  them  be  placed  d  around  in  the  whole  circle,  an  equila-d  L4 
leral   and  equiangular   quindecagon    shall    be    inscribed    in    it- 
Which  was  to  be  done. 


cSO. 


jU  THE  ELEMENTS,  &c. 

Book  IV,  And  in  the  same  manner  as  was  done  in  the  pentagon,  if 
S^^mmJ  through  the  points  of  division  made  by  inscribing  the  quindeca- 
gon,  straight  lines  be  drawn  touching  the  circle,  an  equilateral 
and  equiangular  quindecagon  shall  be  described  about  it :  and 
likewise,  as  in  the  pentagon,  a  circle  may  be  inscribed  in  a  given 
equilateral  and  equiangular  quindecagon,  and  circumscribed 
about  it. 


THE 

ELEMENTS  OF  EUCLID. 


BOOK  V. 


DEFINITIONS. 

I. 

A  LESS  magnitude  is  said  to  be  a  part  of  a  greater  magnitude,  Book  V. 

when  the  less  measures  the  greater,  that  is,  '  when  the  less  is  ^•^f'^ 

'  contained  a  certain  number  of  times  exactly  in  the  greater.' 

II. 

A  greater  magnitude  is  said  to  be  a  multiple  of  a  less,  when  the 
greater  is  measured  by  the  less,  that  is,  '  when  the  greater 
*  contains  the  less  a  certain  number  of  times  exactly.* 

III. 
*  Ratio  is  a  mutual  relation  of  two  magnitudes  of  the  same  kind  S«e  N 
'  to  one  another,  in  respect  of  quantity.' 

IV. 

Magnitudes  are  said  to  have  a  ratio  to  one  another,  when  the 
less  can  be  multiplied  so  as  to  exceed  the  other. 

V. 

The  first  of  four  magnitudes  is  said  to  have  the  same  ratio  to  the  \ 
second,  which  the  third  has  to  the  fourth,  when  any  equimul- 
tiples  whatsoever  of  the  first  and  third  being  taken,  and  any 
equimultiples  whatsoever  of  the  second  and  fourth  ;  if  tlie  mul- 
tiple of  the  first  be  less  than  that  of  the  second,  the  multiple  of 
the  third  is  also  less  than  that  of  the  fourth  ;  or,  if  the  multiple 
of  the  first  be  equal  to  that  of  the  second,  the  multiple  of  the 
third  is  also  equal  to  that  of  the  fourth;  or,  if  the  multiple  of 


120  THE  ELEMENTS 

Book  V.      the  first  be  greater  than  that  of  the  second,  the  multiple  of  the 
•— v^— '      third  is  also  greater  than  that  of  the  fourth. 

VI. 

Magnitudes  which  have  the  same  ratio  are  called  propoi'tionals. 
N.  B.  '  When  four  magnitudes  are  proportionals,  it  is  usually 
*  expressed  by  saying,  the  first  is  to  the  second,  as  the  third  to 
'  the  fourth.' 

VII. 

When  of  the  equimultiples  of  four  magnitudes  (taken  as  in 
the  fifth  definition)  the  multiple  of  the  first  is  greater  than 
that  of  the  second,  but  the  multiple  of  the  third  is  not 
greater  than  the  multiple  of  the  fourth  ;  then  the  first  is  said 
to  have  to  the  second  a  greater  ratio  than  the  third  magni- 
tude has  to  the  fourth ;  and,  on  the  contrary,  the  third  is 
said  to  have  to  the  fourth  a  less  ratio  than  the  first  has  to  the 
second. 

VIII. 

"  Analogy,  or  proportion,  is  the  similitude  of  ratios." 

IX. 

Proportion  consists  in  three  terms  at  least. 

X. 

When  three  magnitudes  are  proportionals,  the  first  is  said  to 
have  to  the  third  the  duplicate  ratio  of  that  which  it  has  to  the 
second. 

XI. 

See  N.  When  four  magnitudes  are  continual  proportionals,  the  first  is 
said  to  have  to  the  fourth  the  triplicate  ratio  of  that  which  it 
has  to  the  second,  and  so  on,  quadruplicate.  Sec  increasing 
the  denomination  still  by  unity,  in  any  number  of  propor- 
tionals. 

Definition  A,  to  wit,  of  compound  ratio. 

When  there  are  any  number  of  magnitudes  of  the  same  kind, 
the  first  is  said  to  have  to  the  hist  of  tliem  the  ratio  com- 
pounded of  the  ratio  which  the  first  has  to  the  second,  and 
of  the  ratio  which  the  second  has  to  the  third,  and  of  the 
ratio  which  the  third  has  to  the  fourth,  and  so  on  unto  the  last 
magnitude. 

Tor  example,  if  A,  B,  C,  D  be  four  magnitudes  of  the  same 
kind,  the  first  A  is  said  to  have  to  the  hist  D  the  ratio  com- 
pounded of  the  ratio  of  A  to  B,  anil  of  the  ratio  of  B  to  C, 
and  of  the  ratio  of  C  to  D  ;  or,  the  ratio  of  A  to  D  is  said  to 
be  compounded  of  the  ratios  of  A  to  B,  B  to  C,  and  C  to  D : 


OF  EUCLID.  121 

Afid  if  A  has  to  B  the  same  ratio  which  E  has  to  F ;  and  B  to  C  Book  V. 
the  same  ratio  that  G  has  to  H ;  and  C  to  D  the  same  that  K  *— y—^ 
has  to  L ;  then,  by  this  definition,  A  is  said  to  have  to  D  the 
ratio  compounded  of  ratios  which  are  the  same  with  the  ratios 
of  E  to  F,  G  to  H,  and  K  to  L  :  and  the  same  thing  is  to  be  un- 
derstood when  it  is  more  briefly  expressed,  by  saying  A  has  to 
D  the  ratio  compounded  of  the  ratios  of  E  to  F,  G  to  H,  and  K 
to  L. 

Jn  like  manner,  the  same  things  being  supposed,  if  M  has  to  N 
the  same  ratio  which  A  has  to  D;  then,  for  shortness'  sake, 
M  is  said  to  have  to  N  the  ratio  compounded  of  the  ratios  of 
E  to  F,  G  to  H,  and  K  to  L. 

XII. 
In  proportionals,  the  antecedent  terms  are  called  homologous  to 

one  another,  as  also  the  consequents  to  one  another. 
*  Geometers  make  use  of  the  following  technical  words  to  sig- 

*  nify  certain  ways  of  changing  either  the  order  or  magnitude 

*  of  proportionals,  so  as  that  they  continue  still  to  be  propor- 

*  tionals.' 

XIII. 
Permutando,   or  alternando,    by   permutation,    or    alternately; 

this  word  is  used  when  there  are  four  proportionals,  and  it  is  See  N- 
inferred,  that  the  first  has  the  same  ratio  to  the  third,  which 
the  second  has  to  the  fourth  ;  or  that  the  first  is  to  the  third,  as 
the  second  to  the  fourth :  as  is  shown  in  the  16th  prop,  of 
this  5 til  book. 

XIV. 
Invertendo,  by  inversion  ;  when  there  are  four  proportionals,  and 
it  is  inferred,  that  the  second  is  to  the  first  as  the  fourth  to  the 
third.    Prop.  B,  book  5. 

XV. 

Componendo,  by  composition  ;  when  there  are  four  proportionals, 
and  it  is  inferred,  that  the  first,  together  with  the  second,  is  to 
the  second,  as  the  third,  together  with  the  fourth,  is  to  the  fourth. 
18  th  prop,  book  5. 

XVI. 
Dividendo,  by  division  ;  when  there  are  four  proportionals,  and  it 
is  inferred,  that  the  excess  of  the  first  above  the  second  is  to 
the  second  as  the  excess  of  the  third  above  the  fourth  is  to  the 
fourth.     17th  prop,  book  5. 

XVII. 

Convertendo,  by  conversion ;  when  there  are  four  proportion- 
als, and  it  is  inferred,  that  the  first  is  to  its  excess  above  the 

Q 


123  THE  ELEMENTS 

BookV.       second,  as  the  third  to  its  excess  above  the  fourth.     Prop.  E^ 
^x-^^^      book  5. 

XVIII. 

Ex  sequali  (sc.  distantia),  or  ex  squo,  from  equality  of  distance  ; 
when  there  is  any  number  of  magnitudes  more  than  two,  and 
as  many  others,  so  that  they  are  proportionals  when  taken 
two  and  two  of  each  rank,  and  it  is  inferred,  that  the  first  is 
to  the  last  of  the  first  rank  of  magnitudes,  as  the  first  is  to  the 
last  of  the  others:  '  Of  this  there  are  the  two  following  kinds, 
*  which  arise  from  the  different  order  in  which  the  magnitudes 
<  are  taken  two  and  two.' 

XIX. 

Ex  aquali,  from  equality ;  this  term  is  used  simply  by  itself, 
when  the  first  magnitude  is  to  the  second  of  the  first  rank» 
as  the  first  to  the  second  of  the  other  rank ;  and  as  the  se- 
cond is  to  the  third  of  the  first  rank,  so  is  the  second  to  the 
third  of  the  other;  and  so  on  in  order,  and  the  inference  is 
as  mentioned  in  the  preceding  definition ;  whence  this  is 
called  ordinate  proportion.  It  is  demonstrated  in  2 2d  prop, 
book  5. 

XX. 

Ex  aequali,  in  proportione  perturbata,  seu  inordinata ;  from  equa- 
lity, in  perturbate  or  disorderly  proportion*  ;  this  term  is  used 
when  the  first  magnitude  is  to  the  second  of  the  first  rank,  as 
the  last  but  one  is  to  the  last  of  the  second  rank ;  and  as  the 
second  is  to  the  third  of  the  first  rank,  so  is  the  last  but  two  to 
the  last  but  one  of  the  second  rank  ;  and  as  the  third  is  to  the 
fourth  of  the  first  rank,  so  is  the  third  from  the  last  to  the  last 
but  two  of  the  second  rank  ;  and  so  on  in  a  cross  order :  and  the 
inference  is  as  in  the  18th  definition.  It  is  demonstrated  in 
the  23d  prop,  of  book  5. 


AXIOMS. 
I. 

EQUIMULTIPLES  of  the  same,  or  ot  equal  magnitudes,  are 
equal  to  one  another. 

*  4  Prop.  lib.  2-  Archimedis  dc  sphsera  et  cylkidro. 


OF  EUCLID. 

II. 

Those  magnitudes  of  which  the  same,  or  equal  magnitudes,  are 
equimultiples,  are  equal  to  one  another. 

III. 

A  multiple  of  a  greater  magnitude  is  greater  than  the  same 
multiple  of  a  less. 

IV. 

That  magnitude  of  which  a  multiple  is  greater  than  the  same 
multiple  of  another,  is  greater  than  that  other  magnitude. 


PROP.  I.     THEOR. 

IF  any  number  of  magnitudes  be  equimultiples  of 
as  many,  each  of  each;  what  multiple  soever  any  one 
of  them  is  of  its  part,  the  same  multiple  shall  all  the 
first  magnitudes  be  of  all  the  other. 

Let  any  number  of  magnitudes,  AB,  CD  be  equimultiples  of 
as  may  others  E,  F,  each  of  each ;  whatsoever  multiple  AB  is 
of  E,  the  same  multiple  shall  AB  and  CD  together  be  of  E  and 
F  together. 

Because  AB  is  the  same  multiple  of  E  that  CD  is  of  F,  as 
many  magnitudes  as  are  in  AB,  equal  to  E,  so  many  are  there 
in  CD,  equal  to  F.      Divide    AB   into   magni- 
tudes equal  to  E,  viz.  AG,  GB  ;   and  CD  into    A  ( 
CH,  HD,  equal  each  of  them  to  F:   the  num-         j 
bcr  therefore  of  the  magnitudes  CH,  HD  shall  | 

be  equal  to  the  number  of  the  others  AG,  GB :     ^T 
and  because    AG    is   equal    to   E,    and    CH   to  j 

F,  therefore  AG  and   CH   together   are   equal     g  I 
to  a  E  and  F  together:    for  the   same   reason,  '  a  As. 2, 

because  GB  is  equal  to  E,  and  FID  to  F;   GB     q  j  i     5 

and  HD  together  are  equal  to  E  and  F  together.  1 

Wherefore,  as  many  magnitudes  as  are  in  AB  I  I 

equal  to  E,   so  many  are  there  in  AB,  CD  to-    H-L 
gether   equal   to    E    and   F    together.      There- 
fore, whatsoever  multiple  AB  is  of  E,  the  same 
multiple  is  AB  and  CD  together  of  E  and  F    ^ 
together. 

Therefore,  if  any  magnitudes,  how  many  soever,  be  equi- 
multiples of  as  many,  each  of  each,  whatsoever  multiple  any 
one  of  them  is  of  its  part,  the  same  multiple  shall  all  the  first 
magnitudes  be  of  all  the  other :  '  For  the  same  demonstration 


!24 


THE  ELEMENTS 


BookV.  '  holds  in  any  numbei-  of  magnitudes,  which  was  here  applied 
•^-nr*-' '  to  two.*    Q.  E.  D. 


PROP.  II.    THEOR. 


IF  the  first  magnitude  be  the  same  multiple  of  the 
second  that  the  third  is  of  the  fourth,  and  the  fifth 
the  same  multiple  of  the  second  that  the  sixth  is  of 
the  fourth ;  then  shall  the  first  together  with  the  fifth 
be  the  same  multiple  of*^*lJ|e  second,  that  the  third 
together  with  the  sixth  is  o^me-i 


-fourth. 


B 


Let  AB  the  first,  be  the  same  multiple  of  C  the  second,  that 
DE  the  third  is  of  F  the  fourth ;  and  BG  the  fifth,  the  same 
multiple  of  C  the   second,   that   EH  j) 

the    sixth   is  of  F    the    fourth  :    then      ^ 
is    AG    the    first,    together    with    the 
fifth,    the    same    multiple    of    C    the 
second,    that    DH  the   third,  together 
with  the  sixth,  is  of  F  the  fourth. 

Because  AB  is  the  same  multiple 
of  C,  that  DE  is  of  F ;  there  are  as 
many  magnitudes  in  AB  equal  to  C, 

as  there  are  in  DE  equal  to  F:  in  like      G  C       H         F 

manner,  as  many  as  there  are  in  BG  equal  to  C,  so  many  are 
there  in  EH  equal  to  F ;  as  many,  then,  as  are  in  the  whole 
AG  equal  to  C,  so  many  are  there  in  the  whole  DH  equal  to 
¥:  therefore  AG  is  the  same  multiple  of  C,  that  DH  is  of  F; 
that  is,  AG  the  first  and  fifth  together,  is 
the  same  multiple  of  the  second  C,  that  D 

DH  the  third  and  sixth  together  is  of  the      A 
fourth  F.      If,  therefore,  the  first  be  the 
same  multiple,  &c.     Q.  E.  D. 

CoR.  '  From  this  it  is  plain,  that,  if  any  B 
'  number  of  magnitudes  AB,  BG,  GH 
'  be  multiples  of  another  C,  and  as  many 
'  DE,  ER,  K.L  be  the  same  multiples  of  G-- 
'  F,  each  of  each,  the  whole  of  the  first, 
*  viz.  AH,  is  the  same  multiple  of  C, 
'  that  the  whole  of  tlie  last,  viz.  DL,  is 
'  of  F.'  H      C        L     F 


K 


OF  EUCLID. 


125 

BookV. 


PROP.  III.    THEOR. 

IF  the  first  be  the  same  multiple  of  the  second, 
which  the  third  is  of  the  fourth;  and  if  of  the  first 
and  third  there  be  taken  equimultiples,  these  shall 
be  equimultiples,  the  one  of  the  second,  and  the 
other  of  the  fourth. 


H 


K 


•  Let  A  the  first,  be  the  same  multiple  of  B  the  second,  that  C 
the  third  is  of  D  the  fourth  ;  and  of  A,  C  let  the  equimultiples 
EF,  GH  be  taken :  then  EF  is  the  same  multiple  of  B,  that  GH 
is  of  D. 

Because  EF  is  the  same  multiple  of  A,  that  GH  is  of  C,  there 
are  as  many  magnitudes  in  EF  equal  to  A,  as  are  in  GH  equal 
to  C  :  let  EF  be  divided  into  the 
magnitudes  EK,  KF,  each  equal 
to  A,  and  GH  into  GL,  LH, 
each  equal  to  C :  the  number 
therefore  of  the  magnitudes  EK, 
KF  shall  be  equal  to  the  number 
of  the  others  GL,  LH :  and  be- 
cause A  is  the  same  multiple  of 
B,  that  C  is  of  D,  and  that  EK  is 
equal  to  A,  and  GL  to  C  ;  there- 
fore EK  is  the  same  multiple  of 
B,  that  GL  is  of  D  ;  for  the  same 
reason,  KF  is  the  same  multiple 
of  B,  that  LH  is  of  D  ;  and  so,  if 
there  be  more  parts  in  EF,  GH 
equal  to  A,  C :  because,  there-  E  A  B 
fore,   the   first  EK  is  the  same 

multiple  of  the  second  B,  which  the  third  GL  is  of  the  fourth 
D,  and  that  the  fifth  KF  is  the  same  multiple  of  the  second  B, 
■which  the  sixth  LH  is  of  the  fourth  D  ;  EF  the  first,  together 
■with  the  fifth,  is  the  same  multiple  »  of  the  second  B,  which  GH  a  2.  5. 
the  third,  together  with  the  sixth,  is  of  the  fourth  D.  If,  there- 
fore, the  first,  &C.    Q.  E.  D. 


C      D 


THE  ELEMENTS 


PROP.  IV.    THEOR. 


See  N.  lY  the  first  of  four  magnitudes  has  the  same  ratio 
to  the  second  which  the  third  hath  to  the  fourth,  then 
any  equimuhiples  whatever  of  the  first  and  third  shall 
have  the  same  ratio  to  any  equimultiples  of  the  second 
and  fourth,  viz.   '  the  equimultiple  of  the  first  shall 

*  have  the  same  ratio  to  that  of  the  second,  which  the 

*  equimultiple  of  the  third  has  to  that  of  the  fourth.' 

Let  A  the  first  have  to  B  the  second  the  same  ratio  which 
the  third  C  has  to  the  fourth  D;  and  of  A  and  C  let  there  be 
taken  any  equimultiples  whatever 
E,  F :  and  of  B  and  D  any  equi- 
multiples whatever  G,  H :  then 
E  has  the  same  ratio  to  G,  which 
F  has  to  H. 

Take  of  E  and  F  any  equimul- 
tiples whatever  K,  L,  and  of  G, 
H,  any  equimultiples  whatever  M, 
N:  then,  because  E  is  the  same 
multiple  of  A,  that  F  is  of  C; 
and  of  E  and  F  have  been  taken 
equimultiples  K,  L  ;  therefore  K 
is  the  same  multiple  of  A,  that  L 

a 3.  5,  isofC*:  for  the  same  reason  M 
is  the  same  multiple  of  B,  that  N 
is  of  D:  and  because    as  A  is  to 

b  Hyp.  E,  so  is  C  to  D^,  and  of  A  and 
C  have  been  taken  certain  equi- 
multiples K,  L;  and  of  B  and  D 
have  been  taken  certain  equimul- 
tiples M,  N ;  if,  ihereforc,  K  be 
greater  than  M,  L  is  greater  than 
N;    and   if  equal,  equal;  if  less, 

c  5  def.5.  less«^.  And  K,  L  are  any  equi- 
multiples whatever  of  E,  F ;  and 
M,  N  any  whatever  of  G,  H : 
as  therefore  E  is  to  G,  so  is'=  F 
to  il.  Therefore,  if  the  first,  Sec. 
Q.  E.  D. 

See  N.  Coil.    Likewise,  if  the  first  has  th.e  same  ratio  to  the  second, 

which  the  third  has  to  (he  fourth,  then  also  any  equimultiples 


K      E 


F 


A 
C 


13 
D 


G 
H 


M 

N 


OF  EUCLID. 


i2r 


whatever  of  the  first  and  third  have  the   same  ratio  to  the  se-  BookV. 
cond  and  fourth :  and  in  like  manner,    the  first  and  the  third  ' — v— ' 
have  the  same  ratio  to  any  equimultiples  whatever  of  the  second 
and  fourth. 

Let  A  the  first  have  to  B  the  second  the  same  ratio  which 
the  third  C  has  to  the  fourth  D,  and  of  A  and  C  let  E  and  Y  he 
any  equimultiples  whatever  ;  then  E  is  to  B,  cvs  F  to  D. 

Take  of  E,  F  any  equimultiples  whatever  K,  L,  and  of  B,  D 
any  equimultiples  whatever  G,  H  ;  then  it  may  be  demonstrated, 
as  before,  that  K  is  the  same  multiple  of  A,  that  L  is  of  C  ;  and 
because  A  is  to  B,  as  C  is  to  D,  and  of  A  and  C  certain  equi- 
multiples have  been  taken,  viz.  K  and  I^ ;  and  of  B  and  D  cer- 
tain equimultiples  G,  H  ;  therefore,  if  K  be  greater  than  G,  L 
is  greater  than  H ;  and  if  equal,  equal;  if  less,  less<^:  and  KcSdef.S 
L  are  any  equimultiples  of  E,  F,  and  G,  H  any  whatever  of  B, 
D  ;  as  therefore  E  is  to  B,  so  is  F  to  D :  and  in  the  same  way 
the  other  case  is  demonstrated. 


PROP.  V.    THEOR* 

IF  one  magnitude  be  the  same  multiple  of  another,  See  n. 
which  a  magnitude  taken  from  the  first  is  of  a  mag- 
nitude taken  from  the  other ;  the  remainder  shall  be 
the  same  multiple  of  the  remainder,  that  the  whole 
is  of  the  whole. 


Let  the  magnitude  AB  be  the  same  multiple      G 
of  CD,  that  AE  taken  from  the  first  is  of  CF 
taken  from  the  other;  the  remainder  EB  shall 
be  the  same  multiple  of  the  remainder  FD,  that      A 
the  whole  AB  is  of  the  whole  CD. 

Take  AG  the  same  multiple  of  FD,  that 
AE  is  of  CF :  therefore  AE  is  ^  the  same  mul- 
tiple of  CF,  that  EG  is  of  CD :  but  AE,  by 
the  hypothesis,  is  the  same  multiple  of  CF  that 
AB  is  of  CD ;  therefore  EG  is  the  same  mul- 
tiple of  CD  that  AB  is  of  CD  ;  wherefore  EG 
is  equal  to  ABb.  Take  from  them  the  common 
magnitude  AE  ;  the  remainder  AG  is  equal  to 
the  remainder  EB.  Wherefore,  since  AE  is 
the  sartie  multiple  of  CF,  that  AG  is  of  FD, 
and  that  AG  is  equal  to  EB  ;  therefore  AE  is  the  same  multiple 
of  CF  that  EB  is  of  FD  :   but  AE  is  ti)e  same  multiple  of  CF 


E— 


B 


al.  5. 


I        bl.Ax.5. 


D 


12B 


THE  ELEMENTS 


BookV.  that  AB  is  of  CD  ;  therefore  EB  is  the  same  multiple  of  FD 
*— y-'*-'  that  AB  is  of  CD.    Therefore,  if  any  magnitude,  Sec    Q.  E.  D. 


PROP.  VI.    THEOR. 


A 


K 


C± 


G~ 


B 


H- 


see  N.  IF  two  magnitudes  be  equimultiples  of  two  others, 
and  if  equimultiples  of  these  be  taken  from  the  first 
two,  the  remainders  are  either  equal  to  these  others, 
or  equimultiples  of  them. 

Let  the  two  magnitudes  AB,  CD  be  equimultiples  of  the  two 

E,  F,  and  AG,  CH  taken  from  the  first  two  be  equimultiples  of 
the  same  E,  F  ;  the  remainders  GB,  HD  are  either  equal  to  E, 

F,  or  equimultiples  of  them. 

First,  Let  GB  be  equal  to  E ;  HD  is 
equal  to  F  :  make  CK  equal  to  F  ;  and 
because  AG  is  the  same  multiple  of  E, 
that  CH  is  of  F,  and  that  GB  is  equal  to 

E,  and  CK  to  F ;  therefore  AB  is  the 
same  multiple  of  E,  that  KH  is  of  F. 
But  AB,  by  the  hypothesis,  is  the  same 
multiple  of  E  that  CD  is  of  F  ;  there- 
fore KH  is  the  same  multiple  of  F,  that 
CD  is  of  F  ;  wherefore  KH  is  equal  to 

al.Ax.5.  CD  a  :  take  away  the  common  magni- 
tude CH,  then  the  remainder  KG  is 

equal  to  the  remainder  HD :  but  KC  is  equal  to  F ;  HD  therefore 

is  equal  to  F. 

But,  let  GB  be  a  multiple  of  E  ;  then 

HD  is  the  same  multiple  of  P':  makeCK 

the  same  multiple  of  F  that  GB  is  of  E: 

and  because  AG  is  the  same  multiple  of 

E  that  CH  is  of  F  ;  and  GB  the  same 

multiple  of  E  that  CK  is  of  F:  therefore 

AB  is  the  same  multiple  of  E  that  KH 
b2.  5.       isofF^:  but  AB  is  the  same  multiple  of 

E  that  CD  is  of  F  ;  therefore  KH  is  the 

same  multiple  of  F  that  CD  is  of  it  ; 

wherefore  KH  is  equal  to  CD^:  take 

away  CH  from  both  ;  therefore  the  re- 
mainder KC  is  equal  to  the  remainder 

HD  :  and  because  GB  is  the  same  multiple  of  E,  that  KC  is  of  F. 

and  that  KC  is  equal  to  HD;  therefore  HD  is  the  same  multiple  of 

F,  that  GB  is  of  E.    If,  therefore,  two  magnitudes,  8cc.    Q.  E.  D» 


D     E      F 


G- 


K 


C— 


H- 


B 


D    E      F 


OF  EUCLID. 


PROP.  A.    THEOR. 

IF  the  first  of  four  magnitudes  has  to  the  second  the  See  n. 
same  ratio  which  the  third  has  to  the  fourth  ;  then,  if 
the  first  be  greater  than  the  second,  the  third  is  also 
greater  than  the  fourth;  and,  if  equal,  equal;  if  less, 
less. 

Take  any  equimultiples  of  each  of  them,  as  the  doubles  of 
each;  then,  by  def.  ^th  of  this  book,  if  the  double  of  the  first  be 
greater  than  the  double  of  the  second,  the  double  of  the  third  is 
greater  than  the  double  of  the  fourth ;  but,  if  the  first  be  greater 
than  the  second,  the  double  of  the  first  is  greater  than  the  double 
of  the  second  ;  wherefore  also  the  double  of  the  third  is  greater 
than  the  double  of  the  fourth  ;  therefore  the  third  is  greater  than 
the  fourth  :  in  like  manner,  if  the  first  be  equal  to  the  second, 
or  less  than  it,  the  third  can  be  proved  to  be  equal  to  the  fourth, 
or  less  than  it.     Therefore,  if  the  first,  &c.     Q.  E.  D. 


PROP.  B.    THEOR. 

IF  four  magnitudes  are  proportionals,    they  are  Se?  i^. 
proportionals  also  when  taken  inversely. 

If  the  magnitude  A  be  to  B  as  C  is  to  D,  then  also  inversely 
B  is  to  A  as  D  to  C. 

Take  of  B  and  D  any  equimultiples 
whatever  E  and  F;  and  of  A  and  C  any 
equimultiples  whatever  G  and  H.  First,  Let 
E  be  greater  than  G,  then  G  is  less  than  E  ; 
and,  because  A  is  to  B  as  C  is  to  D,  and 
of  A  and  C,  the  first  and  third,  G  and  H 
are  equimultiples  ;  and  of  B  and  D,  the  se- 
cond and  fourth,  E  and  F  are  equimulti- 
ples ;  and  that  G  is  less  than  E,  H  is  also  G  A  B  E 
^  less  than  F  ;  that  is,  F  is  greater  than  H  ;  a  5.  defs 

if  therefore  E  be  greater  than  G,  F  is  great-     H      C      D      F  5. 

er  than  H  :  in  like  manner,  if  E  be  equal 
to  G,  F  may  be  shown  to  be  equal  to  H  ; 
and  if  less,  less  ;  and  E,  F  are  any  equi- 
multiples whatever  of  B  and  D,  and  G,  II 
any  whatever  of  A  and  C  ;  therefore,  as  B 

R 


130 


BookV.  is  to  A,    so  is  D   to    C. 
<— v—'  Q.  E.  D. 


THE  ELEMENTS 

If,    then,    four    magnitudes,    &c. 


See  N. 


PROP.  C.    THEOR. 

IF  the  first  be  the  same  muUiple  of  the  second,  or 
the  same  part  of  it,  that  the  third  is  of  the  fourth ; 
the  first  is  to  the  second  as  the  third  is  to  the  fourth. 


A     B     C     D 
E     G      F     H 


Let  the  first  A  be  the  same  multiple  of  B 
the  second,  that  C  the  third  is  of  the  fourth 
D  :  A  is  to  B  as  C  is  to  D. 

Take  of  A  and  C  any  equimultiples  what- 
ever E  and  F  ;  and  of  B  and  U  any  equi- 
multiples whatever  G  and  H  :  then  because 
A  is  the  same  multiple  of  B  that  C  is  of  D  ; 
and  that  E  is  the  same  multiple  of  A  that 
F  is  of  C  ;  E  is  the  same  multiple  of  B  that 

a  3.  5.  F  is  of  D  »  ;  therefore  E  and  F  are  the  same 
multiples  of  B  and  D  :  but  G  and  H  are  equi- 
multiples of  B  and  D ;  therefore,  if  E  be  a 
greater  multiple  of  B  than  G  is,  F  is  a  great- 
er multiple  of  D  than  H  is  of  D  ;  that  is, 
if  E  be  greater  than  G,  F  is  greater  than  H  : 
in  like  manner,  if  E  be  equal  to  G,  or  less, 
F  is  equal  to  H,  or  less  than  it.  But  E,  F 
are  equimultiples,  any  whatever,  of  A,  C, 
and  G,  H  any  equimultiples  whatever  of  B, 

b  S.dci.S.  D.     Therefore  A  is  to  B  as  C  is  to  D  t. 


Next,  Let  the  first  A  be  the  same  part 
of  the  second  B,  that  the  third  C  is  of 
the  fourth  D  :  A  is  to  B  as  C  is  to  D  : 
for  B  is  the  same  multiple  of  A,  that  D 
is  of  C  :  wherefore,  by  the  preceding 
case,  B  is  to  A  as  D  is  to  C ;  and  in- 
c  B  5.  verselyc,  A  is  to  B  as  C  is  to  D.  There- 
fore, if  the  first  be  the  same  multiple, 
8cc.     Q.  E.  D.  A      B      C      D 


OF  EUCLID. 


PROP.  D.     THEOR. 


IF  the  first  be  to  the  second  as  to  the  third  to  the  See  n. 
fourth,  and  if  the  first  be  a  multiple,  or  part  of  the 
second ;  the  third  is  the  same  multiple,  or  the  same 
part  of  the  fourth. 


Let  A  be  to  B  as  C  is  to  D  ;  and  first  let  A  be  a  multiple 
B ;  C  is  the  same  multiple  of  D. 

Take  E  equal  to  A,  and  whatever  mul- 
tiple A  or  E  is  of  B,  make  F  the  same  mul- 
tiple of  D :  then,  because  A  is  to  B  as  C  is 
to  D  ;  and  of  B  the  second  and  D  the  four  h 
equimultiples  have  been  taken  E  and  F  ; 
A  is  to  E  as  C  to  F  a :  but  A  is  equal  to 
E,  therefore  C  is  equal  to  F  •» :  and  F  is 
the  same  multiple  of  D  that  A  is  of  B.  A  B  C  D 
Wherefore  C  is  the  same  multiple  of  D 
that  A  is  of  B. 

Next,  Let  the  first  A  be  a  part  of  the  se- 
cond B  ;  C  the  third  is  the  same  part  of  the 
fourth  D. 

Because  A  is  to  B  as  C  is  to  D  ;  then, 
inversely,  B  is  «  to  A  as  D  to  C:  but  A  is 
'a  part  of  B,  therefore  B  is  a  multiple  of  A; 
and,  by  the  preceding  case,  D  is  the  same 
multiple  of  C,  that  is,  C  is  the  same  part  of 
D,  that  A  is  of  B.    Therefore,  if  the  first,  &c.     Q.  E.  D. 


of 


a  Cor .4.5. 
b  A.5. 


See  the 
figure  at 
the  foot 
of  the 
preced- 
ing pag<^ 
cB.  5. 


PROP.  Vn.    THEOR. 


EQUAL  magnitudes  have  the  same  ratio  to  thje 
same  magnitude ;  and  the  same  has  the  same  ratio  to 
equal  magnitudes. 

Let  A  and  B  be  equal  magnitudes,  and  C  any  other.  A  and 
B  have  each  of  them  the  same  ratio  to  C,  and  C  has  the  same 
ratio  to  each  of  the  magnitudes  A  and  B. 

Take  of  A  and  B  any  equimultiples  whatever  D  and  E,  and 


132 


THE   ELEMENTS 


BookV   of  C  any  multiple  whatever  F:    then,  because  D  is  the  same 
''—V—'  n)ultiple  of  A  that  E  is  of  B,  and  that  A  is 
a  l.Ax.5.  equal  to  B  ;  D  is  ^  equal  to  E  :  therefore,  if 

D  be  greater  than  F,  E  is  greater  than  F  ; 

and  if  equal,  equal ;  if  less,  less :  and  D,  E 

are  any  equimultiples  of  A,  B,  and  F  is  any 
b5.def.5.  multiple  of  C.      Therefore  b,  as  A  is  to  C, 

so  is  B  to  C. 

Likewise  C  has  the  same  ratio  to  A,  that 

it  has  to  B  :   for,  having  made  the  same  con- 
struction, D  may  in  like  manner  be  shown 

equal  to  E :  therefore,  if  F  be  greater  than 

D,  it  is  likewise  greater  than  E  ;  and  if  equal, 

equal ;  if  less,  less  :  and  F  is  any  multiple 

whatever  of  C,  and  D,  E  are  any  equimulti- 
ples whatever  of  A,  B.    Therefore  C  is  to  A 

as  C  is  to  B''.  Therefore,  equal  magnitudes, 

&c.     Q.  E.  D. 


See  N. 


D 

E 


A 
B 


PROP.  Vin.     THEOR. 

OF  unequal  magnitudes,  the  greater  has  a  greater 
ratio  to  the  same  than  the  less  has ;  and  the  same 
magnitude  has  a  greater  ratio  to  the  less,  than  it  has 
to  the  greater. 

Let  AB,   BC  be  unequal   magnitudes,    of   which  AB  is  the 
greater,    and    let    D    be    any  magnitude 
whatever  :   AB  has  a  greater  ratio  to  D  Fig.  1. 

than  BC  to  D:  and  D  has  a  greater  ra- 
tio to  BC  than  unto  AB. 

If  the    magnitude    which    is   not   the 
greater  of  the  two  AC,  CB,  be  not  less  A 

than  D,  lake  EF,  EG,  the  doubles  of 
AC,  CB,  as  in  Fig.  1.  But,  if  that  which 
is  not  the  greater  of  the  two  AC,  CB 
be  less  than  D  (as  in  Fig.  2.  and  3.)  this 
magnitude  can   be  multiplied,   so   as  to  G      B 

become   greater  than  D,   whether  it  be 

AC,  or  CB.     Let  it  be  multiplied  until  L      K      H      D 

it  become  greater  than  D,  and  let  the 
other  be  multiplied  as  often;  and  Itt  EF 
be  the  multiple  thus  taken  of  AC,  and 
FG  the  same  multiple  of  CB :  therefore 
EF  and  FG  are  each  of  them  greater  than 


F  — 


OF  EUCLID. 


1J3 


D :  and  in  every  one  of  the  cases,  take  H  the  double  of  D,  K  BookV. 
its  triple,  and  so  on,  till  the  multiple  of  D  be  that  which  first  •*— #^-^ 
becomes  greater  than  FG :  let  L  be  that  multiple  of  D  which  is 
first  greater  than  FG,  and  K  the  multiple  of  D  which  is  next 
less  than  L. 

Then,  because  L  is  the  multiple  of  D,  which  is  the  first  that 
becomes  greater  than  FG,  the  next  preceding  multiple  K  is 
not  greater  than  FG ;  that  is,  FG  is  not  less  than  K :  arwl  since 
EF  is  the  same  multiple  of  AC,  that  FG  is  pf  CB ;  FG  is  the 
same  multiple  of  CB,  that  EG  is  of  AB^;  wherefore  EG  and»l*5- 
FG  are  equimultiples  of  AB  and  CB ;  and  it  was  shown,  that 
FG  was  not  less  than 


Fig.  2, 


Fig.  3. 


G 


A 
C- 


B 


F— 


G 


K,  and,  by  the  con- 
struction, EF  is  great- 
er than  D ;  therefore 
the  whole  EG  is  great- 
er than  K  and  D  toge- 
ther :  but  K,  together 
with  D,  is  equal  to  L  ; 
therefore  EG  is  great- 
er than  L ;  but  FG  is 
not  greater  than  L ; 
and  EG,  FG  are  equi- 
multiples of  AB,  BC, 
and  L  is  a  multiple  of 
D  ;  therefore^  AB  has 
to  D  a  greater  ratio 
than  BC  has  to  D. 

Also,  D  has  to  BC  a 
greater  ratio  than  it 
has  to  AB :  for,  hav- 
ing made  the  same 
construction,  it  may 
be  shown,  in  like  man- 
ner, that  L  is  greater 
than  FG,  but  that  it  is  not  greater  than  EG :  and  L  is  a  multiple 
ofD;  and  FG,  EG  are  equimultiples  of  CB,  AB :  therefore  D 
has  to  CB  a  greater  ratio''  than  it  has  to  AB.  Wherefore,  of 
unequal  magnitudes,  &c.     Q.  E.  D. 


K     H     D 


C-- 
B 


K     D 


b  7.  def. 
5. 


134 


THE  ELEMENTS 


Book  V. 


PROP.  IX.     THEOR. 


See  N.  MAGNITUDES  which  have  the  same  ratio  to  the 
same  magnitude  are  equal  to  one  another;  and  those 
to  which  the  same  magnitude  has  the  same  ratio  are 
equal  to  one  another. 


Let  A,  B  have  each  of  them  the  same  ratio  to  C:  A  is  equal 
to  B:  for,  if  they  are  not  equal,  one  of  them  is  greater  than  the 
other;  let  A  be  the  greater;  then,  by  what  was  shown  in  the 
preceding  proposition,  there  are  some  equimultiples  of  A  and  B, 
and  some  multiple  of  C  such,  that  the  multiple  of  A  is  greater 
than  the  multiple  of  C,  but  the  multiple  of  B  is  not  greater  than 
that  of  C.  Let  such  multiples  be  taken,  and  let  13,  E  be  the 
equimultiples  of  A,  B,  and  F  the  multiple  of  C,  so  that  D  may 
be  greater  than  F,  and  E  not  greater  than  F :  but,  because  A  is 
to  C  as  B  is  to  C,  and  of  A,  B  are  taken 
equimultiples  D,  E,  and  of  C  is  taken  a 
multiple  F  ;  and  that  D  is  greater  than  F  ; 
a5.  def.   E  shall  also  be  greater  than  F^;  but  E  is  D 

^'  not  greater  than  F,   which  is  impossible; 

A  therefore  and  B  are  not  unequal ;  that  is, 
they  are  equal. 

Next,  Let  C  have  the  same  ratio  to  each 
of  the  magnitudes  A  and  B ;  A  is  equal  to 
B :  for,  if  they  are  not,  one  of  them  is 
greater  than  the  other;  let  A  be  the 
greater;  therefore,  as  was  shown  in  Prop. 
8lh,  there  is  some  multiple  F  of  C,  and 
some  equimultiples  E  and  D,  of  B  and  A 
such,  that  F  is  greater  than  E,  and  not  greater  than  D  ;  but  be- 
cause C  is  to  B,  as  C  is  to  A,  and  that  F,  the  multiple  of  the 
first,  is  greater  than  E,  the  multiple  of  the  second ;  F,  the  mul- 
tiple of  the  third,  is  greater  than  D,  the  multiple  of  the  fourth*: 
but  F  is  not  greater  than  D,  which  is  impossible.  Therefore  A 
is  equal  to  B.     Wherefore,  magnitudes  which,  &c.     Q.  E.  D. 


A 


B 


OF  EUCLID. 


PROP.  X.    THEOR. 


THAT  magnitude  which  has  a  greater  ratio  than  See  N. 
another  has  unto  the  same  magnitude  is  the  greater 
of  the  two :  and  that  magnitude,  to  which  the  same 
has  a  greater  ratio  than  it  has  unto  another  magnitude, 
is  the  lesser  of  the  two. 


Let  A  have  to  C  a  greater  ratio  than  B  has  to  C :  A  is  great- 
er than  B:     for,  because  A  has   a  greater  ratio  to  C  than  B 
has  to  C,    there   are  *  some    equimultiples  of  A   and  B,    and  a  7- def 
some  multiple  of  C  such  that  the  multiple  of  A  is  greater  than     5. 
the  multiple  of  C,  but  the  multiple  of  B  is  not  greater  than  it: 
let  them  be  taken,  and  let  D,  E  be  equi- 
multiples of  A,  B,  and  F  a  multiple  of  C 
such  that  D  is  greater  than   F,  but  E  is 
not  greater  than  F :  therefore  D  is  greater 
than  E:    and,  because  D  and  E  are  equi-      A  D 

multiples  of  A   and  B,   and  D   is  greater 
than  E  ;  therefore  A  is  *»  greater  than  B.  I  l*'      b  4.  Ak. 

Next,   Let  C  have  a  greater  ratio  to  B  CI  5. 

than  it  has  to  A ;  B  is  less  than  A:  for* 
there  is  some  multiple  F  of  C,  and  some 
equimultiples  E  and  D  of  B  and  A  such  B 
that  F  is  greater  than  E,  but  is  not  greater 
than  D :  E  therefore  is  less  than  D  ;  and 
because  E  and  D  are  equimultiples  of  B 
and  A,  therefore  B  is  ''  less  than  A.  That 
magnitude,  therefore,  &;c.     Q.  E.  D. 


PROP.  XL    THEOR. 

RATIOS  that  are  the  same  to  the  same  ratio  are 
the  same  to  one  another. 

Let  A  be  to  B  as  C  is  to  D ;  and,  as  C  to  D,  so  let  E  be  to 
F;  A  is  to  B  as  E  to  F. 

Take  of  A,  C,  E  any  equimultiples  whatever  G,  H,  K ;  and 
of  B,  D,  F  any  equimultiples  whatever  L,  M,  N.  Therefore, 
since  A  is  to  B  as  C  to  D,  and  G,  H  are  taken  equimultiples  of 


136  THE  ELEMENTS 

Book  V.  A,  C,  and  L,  M  of  B,  D  ;  if  G  be  greater  than  L,  H  is  greater 
than  M ;  and  if  equal,  equal ;  and  if  less,  less  *.  Again,  be- 
cause C  is  to  D  as  E  is  to  F,  and  H,  K  are  taken  equimultiples 
of  C,  E ;  and  M,  N  of  D,  F :  if  H  be  greater  than  M,  K  is 
greater  than  N  j  and  if  equal,  equal ;   and  if  less,  less:    but  if  G 


H- K- 

C E- 


B D- 


L M- 


be  greater  than  L,  it  has  been  shown  that  H  is  greater  than  M ; 
and  if  equal,  equal;  and  if  less,  less;  therefore,  if  G  be  greater 
than  L,  K  is  greater  than  N;  and  if  equal,  equal;  and  if  less, 
less:  and  G,  K  are  any  equimultiples  whatever  of  A,  E;  and 
L,  N  any  whatever  of  B,  F :  therefore,  as  A  is  to  B  so  is  E  to 
F».     Wherefore,  ratios  that,  &c.     Q.  E.  D. 


PROP.  XII.     THEOR. 

IF  any  number  of  magnitudes  be  proportionals,  as 
one  of  the  antecedents  is  to  its  consequent,  so  shall 
all  the  antecedents  taken  together  be  to  all  the  conse- 
quents. 

Eet  any  number  of  magnitudes  A,  B,  C,  D,  E,  F  be  propor- 
tionals ;  that  is,  as  A  is  to  B  so  C  to  D,  and  E  to  F :  as  A  is 
to  B,  so  shall  A,  C,  E  together  be  to  B,  D,  F  together. 

Take  of  A,  C,    E  any   equimultiples    whatever    G,    H,    K ; 


H K- 


A C E- 

B D F- 


•M N- 


and  of  B,  D,  F  any  equimultiples  whatever  L,  M,  N:    then, 
because  A  is  to  B  as  C  is  to  D,  and  as  E  to  F ;   and  that  G,  H, 


OF  EUCLID.  13? 

K  are  equimultiples  of  A,  C,  E,  and  L,  M,  N  equimultiples  of  BookV. 

B,  D,  F ;  if  G  be  greater  than  L,  H  is  greater  than  M,  and  K  ^— v».»* 
greater  than  N ;  and  if  equal,  equal ;  and  if  less,  less*.  Where-a5.  def.5. 
fore,  if  G  be  greater  than  L,  then  G,  H,  K  together  are  greater 

than  L,  M,  N  together ;  and  if  equal,  equal ;  and  if  less,  less. 
And  G,  and  G,  H,  K  together  are  any  equimultiples  of  A,  and 
A,  C,  E  together;  because,  if  there  be  any  number  of  magni-  * 
tudes  equimultiples  of  as  many,  each  of  each,  whatever  multi- 
ple one  of  them  is  of  its  part,  the  same  multiple  is  the  whole  of 
the  whole  •> :  for  the  same  reason  L,  and  L,  M,  N  are  any  equi-  b  1.  5. 
multiples  of  B,  and  B,  D,  F :  as  therefore  A  is  to  B,  so  are  A, 

C,  E  together  to  B,  D,  F  together.     Wherefore,  if  any  number, 
&c.     Q.  E.  D. 


PROP.  XIII.    THEOR. 

IF  the  first  has  to  the  second  the  same  ratio  which  See  N. 
the  third  has  to  the  fourth,  but  the  third  to  the  fourth 
a  greater  ratio  than  the  fifth  has  to  the  sixth;  the 
first  shall  also  have  to  the  second  a  greater  ratio  than 
the  fifth  has  to  the  sixth. 

Let  A  the  first  have  the  same  ratio  to  B  the  second,  which  C 
the  third  has  to  D  the  fourth,  but  C  the  third  to  D  the  fourth, 
a  greater  ratio  than  E  the  fifth  to  F  the  sixth :  also,  the  first  A 
shall  have  to  the  second  B,  a  greater  ratio  than  the  fifth  E  to  the 
sixth  F. 

Because  C  has  a  greater  ratio  to  D,  than  E  to  F,  there  are 
some  equimultiples  of  C  and  E,  and  some  of  D  and  F  such, 
that   the  multiple  of  C  is  greater  than  the  multiple  of  D,  but 

M. . G H 

A .  C— E 

B D F 

N K L 


the  multiple  of  E  is  not  greater  than  the  multiple  of  F^:  Ieta7.def.5. 
such  be  taken,  and  of  C,  E  let  G,  H  be  equimultiples,  and  K,  L 
equimultiples  of  D,  F,  so  that  G  be  greater  than  K,  but  H  not 
greater  than  L ;  and  whatever  multiple  G  is  of  C,  take  M  the 
same  multiple  of  A  ;  and  whatever  multiple  K  is  of  D,  take  N  the 
same  multiple  of  B :  then,  because  A  is  to  B,  as  C  to  D,  and 

S 


138 


THE  ELEMENTS 


Book  V.  of  A  and  C,  M  and  G  are  equimultiples ;  and  of  B  and  D,  N 

^-"ymmJ  and  K  are  equimultiples ;  if  M  be  greater  than  N,  G  is  greater 

b  5.def.5.  than  K;  and  if  equal,  equal;  and  if  less,  less^;  but  G  is  greater 

than  K,  therefore  M  is  greater  than  N :  but  H  is  not  greater  than 

L ;  and  M,  H  are  equimultiples  of  A,  E  ;  and  N,  L  equimultiples 

c7.def.5.  of  B,   F :  therefore  A  has  a  greater  ratio  to  B  than  E  has  to  F<=. 

Wherefore,  if  the  first,  8cc.     Q.  E.  D. 

Cor.  And  if  the  first  has  a  greater  ratio  to  the  second,  than 
the  third  has  to  the  fourth,  but  the  third  the  same  ratio  to  the 
fourth,  which  the  fifth  has  to  the  sixth  ;  it  may  be  demonstrated, 
in  like  manner,  that  the  first  has  a  greater  ratio  to  the  second, 
than  the  fifth  has  to  the  sixth. 


PROP.  XIV.     THEOR. 


SeeN. 


A  8.5. 


b  13.  5. 
c  10.  5. 


d9.5. 


IF  the  first  has  to  the  second  the  same  ratio  which 
the  third  has  to  the  fourth;  then,  if  the  first  be  great- 
er than  the  third,  the  second  shall  be  greater  than  the 
fourth;  and  if  equal,  equal;  and  if  less,  less. 

Let  the  first  A  have  to  the  second  B,  the  same  ratio  which 
the  third  C  has  to  the  fourth  D  5  if  A  be  greater  than  C,  B  is 
greater  than  D. 

Because  A  is  greater  than  C,  and  B  is  any  other  magnitude, 
A  has  to  B  a  greater  ratio  than  C  to  B*:  but,  as  A  is  to  B,  so 


A  B  C  D 


A   B  C  D 


A  B   C   D 


is  C  to  D  ;  therefore  also  C  has  to  D  a  greater  ratio  than  C  has 
to  B'':  but  of  two  magnitudes,  that  to  which  the  same  has  the 
greater  ratio  is  the  lesser  <= :  wherefore  D  is  less  than  B  ;  that  is, 
B  is  greater  than  D. 

Secondly,  If  A  be  equal  to  C,  B  is  equal  to  D :  for  A  is  to  B,  as 
C,  that  is.  A,  to  D ;  B  therefore  is  equal  to  D^. 

Thirdly,  If  A  be  less  than  C,  B  shall  be  less  than  D :  for  C  is. 
greater  than  A,  and  because  C  is  to  D,  as  A  is  to  B,  D  is  greater 
than  B,  by  the  first  case ;  wherefore  B  is  less  than  D.  There- 
fore, if  the  first,  &c.     Q.  E.  D. 


OF  EUCLID. 


PROP.  XV.  THEOR. 


MAGNITUDES  have*  the  same  ratio  to  one  an- 
other  which  their  equimultiples  have. 


Let  AB  be  the  same  multiple  of  C  that  DE  is  of  F :  C  is  to  F 
as  AB  to  DE. 

Because  AB  is  the  same  multiple  of  C  that  DE  is  of  F,  there 
are  as  many  magnitudes  in  AB  equal  to  C      ^ 
as  there  are  in  DE  equal  to  F :   let  AB  be 
divided  into  magnitudes,  each  equal  to  C, 
viz.  AG,  GH,  HB ;    and  DE  into  magni- 
tudes, each  equal  to  F,  viz.  DK,  KL,  LE ;  G 
then  the  number  of  the  first  AG,  GH,  HB 
shall  be  equal  to  the  number  of  the  last  DK, 
KL,  LE :    and  because  AG,  GH,  HB  are  H 
ail  equal,  and  that  DK,  KL,  LE  are  also 
equal  to  one  another :    therefore  AG  is  to 
DK  as  GH  to  KL,  and  as  HB  to  LE=^:  I       ^7.5. 

and  as  one  of  the  antecedents  to  its  conse-       B      C      E      F 
quent,  so  are  all  the  antecedents  together  to  all  the  consequents 
together'^;    Avherefore,  as  AG  is  to  DK  so  is  AB  to  DE :    but  b  12 
AG  is  equal  to  C,  and  DK  to  F :    therefore,  as  C  is  to  F  so  is 
AB  to  DE.     Therefore,  magnitudes,  &c.     Q.  E.  D. 


D 


K— 


PROP.  XVL    THEOR. 


IF  four  magnitudes  of  the  same  kind  be  propor- 
tionals, they  shall  also  be  proportionals  when  taken 
alternately. 

Let  the  four  magnitudes  A,  B,  C,  D  be  proportionals,  viz.  as 
A  to  B  so  C  to  D ;  they  shall  also  be  proportionals  Avhen  taken 
alternately,  that  is,  A  is  to  C  as  B  to  D. 

Take  of  A  and  B  any  equimultiples  whatever  E  and  F ;  and 
of  C  and  D  take  any  equimultiples  whatever  G  and  H:    and 


140  THE  ELEMENTS 

Book  V.  because  E  is  the  same  multiple  of  A  that  F  is  of  B,    and  that 
•"— v-—*  magnitudes  have   the   same   ratio  to  one   another   which   their 
a  15.  5.     equimultiples  have  a;  therefore  A  is  to  B,   as  E  is  to  F  :  but  as 
A  is  to  B,  so  is  C  to  D  : 

wherefore,  as  C  is  to  D,     E G 

b  11.  5.    so  t*  is  E  to  F:  again,  be- 
cause G,  H  are  equimul-     A C — ■ 

tiples  of  C,  D,  as  C  is  to 


D,  so  is  G  to  II  a ;  but    B- D 

as  C  is  to  D,  so  is  E  to 

F.     Wherefore,  as  E  is     F —     H 

to  F,  so  is  G    to   HW 

But,  when   four  magnitudes   are   proportionals,   if  the  first  be 

greater  than  the  third,  the  second  shall  be  greater  than  the 
c  14.  5.     fourth;  and  if  equal,  equal ;  if  less,  less  c.     Wherefore,  if  E  be 

greater  than  G,  F  likewise  is  greater  than  H  ;  and  if  equal,  equal ; 

if  less,  less  :  and  E,  F  are  any  equimultiples  whatever  of  A,  B  ;  and 
d5.def.5.  G,  H  any  whatever  of  C,  D.     Therefore,  A  is  to  C,  as  B  to  D^ 

If  then  four  magnitudes,  8cc.     Q.  E.  D, 


PROP.  XVII.    THEOR. 


SeeN.  IF  magnitudes,  taken  jointly,  be  proportionals, 
they  shall  also  be  proportionals  when  taken  sepa- 
rately; that  is,  if  two  magnitudes  together  have  to 
one  of  them  the  same  ratio  which  two  others  have  to 
one  of  these,  the  remaining  one  of  the  first  two  shall 
have  to  the  other  the  same  ratio  which  the  remaining 
one  of  the  last  two  has  to  the  other  of  these. 

Let  AB,  BE,  CD,  DF  be  the  magnitudes  taken  jointly  which 
are  proporitonals ;  that  is,  as  AB  to  BE,  so  is  CD  to  DF ;  they 
shall  also  be  proportionals  taken  separately,  viz.  as  AE  to  EB,  so 
CF  to  FD. 

Take  of  AE,  EB,  CF,  FD  any  equimultiples  whatever  GH, 
IlK,  LM,  MN ;  and  again,  of  EB,  FD  take  any  e<juimultiples 
w  hatever  KX,  NP :  and  because  GH  is  the  same  multiple  of 
a  1.5.  AE,  that  HK  is  of  EB,  wherefore  GH  is  the  same  multiple*  of 
AE,  t!iat  GK  is  of  AB :  but  GH  is  the  same  multiple  of  AE, 
that  LM  is  of  CF ;  wherefore  GK  is  the  same  multiple  of  AB, 


OF  EUCLID. 


141 


thafLM  is  of  CF.     Again,  because  LM  is  the  same  multiple  of  BookV. 
CF,  that  MN  is  of  FD  ;  therefore  LM  is  the  same  muUiple  »  of  ^-— y^^iJ 
CF,  that  LN  is  of  CD  :  but  LM   was   shown  to  be   the  same 
multiple  of  CF,  that  GK  is  of  AB  ;  GK  therefore  is  the  same 
multiple  of  AB,  that  LN  is  of  CD  ;  that  is,  GK,  LN  are  equi- 
multiples of  AB,  CD.     Next,  because  HK  is  the  same  multiple 


X 


K— 


H  — 


B 


of  EB,  that  MN  is  of  FD  ;  and  that  KX  is 
also  the  same  multiple  of  EB,  that  NP  is 
of  FD  ;  therefore  HX  is  the  same  multiple 
b  of  EB,  that  MP  is  of  FD.  And  because 
AB  is  to  BE,  as  CD  is  to  DF,  and  that  of 
AB  and  CD,  GK  and  LN  are  equimulti- 
ples, and  of  EB  and  FD,  HX  and  MP  are 
equimultiples  ;  if  GK  be  greater  than  HX,  jg- 

then  LN  is  greater  than  MP  ;  and  if  equal, 
equal;  and  if  less,  less  <= :  but  if  GH  be 
greater  than  KX,  by  adding  the  common 
part  HK  to  both,  GK  is  greater  than  HX  ; 
%vherefore  also  LN  is  greater  than  MP;  j7__ 

and  by  taking  away  MN  from  both,  LM 
is  greater  than  NP  :  therefore,  if  GH  be 
greater  than  KX,  LM  is  greater  than  NP. 
In  like  manner  it  may  be  demonstrated,  G  A  C 
that  if  GH  be  equal  to  KX,  LM  likewise  is  equal  to  NP  ;  and 
if  less,  less  :  and  GH,  LM  are  any  equimultiples  whatever  of 
AE,  CF,  and  KX,  NP  are  any  whatever  of  EB,  FD.  There- 
fore S  as  AE  is  to  EB  so  is  CF  to  FD.  If,  then,  magnitudes, 
8cc.    Q.  E.  D. 


b2.  5. 


D 


M 


c5.dcf.5. 


PROP.  XVIII.     THEOR. 

IF  magnitudes,  taken  separately,  be  proportionals,  see  n. 
they  shall  also  be  proportionals  when  taken  jointly, 
that  is,  if  the  first  be  to  the  second,  as  the  third  to 
the  fourth,  the  first  and  second  together  shall  be  to 
the  second,  as  the  third  and  fourth  together  to  the 
fourth. 


Let  AE,  EB,  CF,  FD  be  proportionals  ;  that  is,  as  AE  to 
EB,  so  is  CF  to  FD  ;  they  shall  also  be  piopovtionals  when  ta- 
ken jointly  ;  that  is,  as  AB  to  BE,  so  CD  to  DF. 

Take  of  AB,  BE,  CD,  DF  any  equimultiples  whatever  GH, 
HK,  LM,  MN  :  and  again,  of  BE,  DF  take  any  whatever  equi- 


14!^ 


THE  ELEMENTS 


Book  V.  multiples  KO,  NP :  and  because   KO,  NP  are    equimultiples 
^— v^*-*  of  BE,  DF;  and  that  KH,  NM  are  equimultiples   likewise  of 

BE,  DF,  if  KO,  the  multiple  of  BE,  be  greater  than  KH,  which 

is  a  multiple  of  the  same  BE,  NP,  likewise  the  multiple  of  DF, 

shall  be  greater  than  NM,  the  multiple 

of  the  same  DF  ;  and  if  KO  be  equal  H 

to  KH,  NP  shall  be  equal  to  NM ;   and 

if  less,  less.  O" 

First,  let  KO  not  be  greater  than  KH, 

therefore  NP  is  not  greater  than  NM  : 

and  because  GH,  HK  are  equimultiples 

of  AB,  BE,  and  that  AB  is  greater  than   K- 
a3.Ax.5.  BE^  therefore  GH  is  greater  »  than  HK  ; 

but  KO  is  not  greater  than  KH,  where- 
fore GH  is  greater  than  KO.     In  like 

manner  it  may  be  shown,  that  LM  is 

greater  than  NP.     Therefore,  if  KO  be 

not    t^reater    than    KH,    then  GH,  the 

multiple  of  AB,  is  always  greater  than 

KO,  the  multiple  of  BE;  and  likewise 

LM,  the  multiple  of  CD,  greater  than  ^ 

NP,  the  multiple  of  DF. 


M 
P-- 

N 


D 

f| 

C       L 


H 


K  — 


M 


N— 


Next,  Let  KO  be  greater  than  KH:  therefore,  as  has  been 
shown,  NP  is  greater  than  NM :  and  because  the  whole  GH  is 
the  same  multiple  of  the  Avhole  AB,  that  HK  is  of  BE,  the  re- 
mainder GK  is  the   same    multiple   of  q  i 

hS.5.  the  remainder  AE  that  GH  is  of  AB^*:  | 
which  is  the  same  that  LM  is  of  CD. 
In  like  manner,  because  LM  is  the 
same  multiple  of  CD,  that  MN  is  of 
DF,  the  remainder  LN  is  the  same 
multiple  of  the  remainder  CF,  that 
the  whole  LM  is  of  the  whole  CD  •> : 
but  it  was  shown  that  LM  is  the  same 
multiple  of  CD,  that  GK  is  of  AE ; 
therefore  GK  is  the  same  multiple  of 
AE,  that  LN  is  of  CF ;  that  is,  GK,  E 

LN   are   equimultiples   of    AE,     CF : 
and   because    KO,    NP    are    equimul- 
tiples  of  BE,   DF,   if  from   KO,   NP   G 
ti.ere  be    taken  KH,    NM,    which    are   likewise   equimultiples 
of  BE,  DF,  the  remainders  HO,  MP  are  either  equal   to  BE, 

r  f>.  5.      i^F)   or   equimultiples   of    lhem<:.      First,    Let    HO,    MP     be 
L-qual  to  BI;.  DF ;  and  because  AE  is  lo  EB^  as.  CF  to  FD,  and 


B 


D 
C       L 


OF  EUCLID. 


143 


that  GK,  LN  are  equimultiples  of  AE,  CF ;  GK  shall  be  to  BookV 
EB,  as  LN  to  FD  d :  but  HO  is  equal  to  EB,  and  MP  to  FD  ;  <>.-v^ 
wherefore  GK  is  to  HO  as  LN  to  MP.     If,  therefore,  GK  bedCor.4.5. 
greater  than  HO,  LN  is  greater  than  MP  ;  and  if  equal,  equal; 
and  if  less  «,  less.  e  A.  5. 

But  let  HO,  MP  be  equimultiples  of  EB,  FD  ;  and  because 
AE  is  to  EB  as  CF  to  FD,  and  that  of  AE,  CF  are  taken  equi- 
multiples GK,  LN  ;  and  of  EB,  FD,  the  equimultiples  HO, 
MP  ;  if  GK  be  greater  than  HO,  LN 
is  greater  than  MP ;    and  if  equal, 

equal ;  and  if  less,  less  ^  ;  which  was  f  5.def.5. 

likewise  shown  in  the  preceding 
case.  If,  therefore,  GH  be  greater 
than  KO,  taking  KH  from  both,  GK 
is  greater  than  HO  ;  wherefore  also 
LN  is  greater  than  MP  ;  and,  conse- 
quently, adding  NM  to  both,  LM  is 
greater  than  NP  :  therefore,  if  GH 
be  greater  than  KO,  LM  is  great- 
er than  NP.  In  like  manner  it 
may  be  shown,  that  if  GH  be  equal 
to  KO,  LM  is  equal  to  NP ;  and  if 
less,  less.  And  in  the  case  in  which 
KO  is  not  greater  than  KH,  it  has 

been  shown  that  GH  is  always  greater  than  KO,  and  likewise 
LM  than  NP  :  but  GH,  LM  are  any  equimultiples  of  AB,  CD, 
and  KO,  NP  are  any  whatever  of  BE,  DF  ;  therefore  f,  as  AB  is 
to  BE  so  is  CD  to  DF.     If  then  magnitudes,  &c.     Q.  E.  D. 


O 


H— 


K  — 


M- 


N— 


B 


E— 


D 

F-- 


PROP.  XIX.    THEOR. 


IF  a  whole  magnitude  be  to  a  whole,  as  a  magnitude  See  n. 
taken  from  the  first  is  to  a  magnitude  taken  from 
the  other  ;  the  remainder  shall  be  to  the  remainder, 
as  the  whole  to  the  whole. 


Let  the  whole  AB  be  to  the  whole  CD  as  AE,  a  magnitude 
taken  from  AB,  to  CF,  a  magnitude  taken  from  CD  ;  the  re- 
manider  EB  shall  be  to  the  remainder  FD  as  the  whole  AB  to 
the  whole  CD. 

Because  AB  is  to  CD  as  AE  to  CF  ;  likewise,  alternately  »,  a  16.  S. 


144 


THE  ELEMENTS 


E— 


BookV.  BA  is  to  AE  as  DC  to  CF  :  and  because,  if  mag- 
v,..y...^  nitudes,  taken  jointly,  be  proportionals,  they  are 
b  17.  5.  also  proportionals  ^  when  taken  separately  ;  there- 
fore, as  BE  is  to  DF  so  is  EA  to  FC  ;  and  alter- 
nately, as  BE  is  to  EA,  so  is  DF  to  FC :  but,  as 
AE  to  CF,  so  by  the  hypothesis  is  AB  to  CD  ; 
therefore  also  BE,  the  remainder,  shall  be  to  the 
remainder  DF,  as  the  whole  AB  to  the  whole 
CD.     Wherefore,  if  the  whole,  Sec.     Q.  E.  D. 

Cor.  If  the  whole  be  to  the  whole,  as  a  mag-  " 
nitude  taken  from  the  first  is  to  a  magnitude  taken 
from  the  other  ;  the  remainder  likewise  is  to  the 
remainder,  as  the  magnitude  taken  from  the  first  to  that  taken 
from  the  other  ;  the  demonstration  is  contained  in  the  pi*eceding. 


B 


D 


PROP.  E.     THEOR. 


IF  four  magnitudes  be  proportionals,  they  are  also 
proportionals  by  conversion,  that  is,  the  first  is  to  its 
excess  above  the  second,  as  the  third  to  its  excess 
above  the  fourth. 


a  17. 5. 
bB.  5. 
c  18.  5. 


Let  AB  be  to  BE  as  CD  to  DF  ;  then  BA  is 
to  AE  as  DC  to  CF. 

Because  AB  is  to  BE  as  CD  to  DF,  by  divi- 
sion a,  AE  is  to  EB  as  CF  to  FD  ;  and  by  inver- 
sion b,  BE  is  to  E  A  as  DF  to  FC.     Wherefore,  by 


composition  S  BA  is  to  AE,  as  DC  is  to  CF. 
therefore,  four,  Sec.     Q.  E.  D. 


If, 


B 


F— 


D 


PROP.  XX.     THEOR. 


SeeN. 


IF  there  be  three  magnitudes,  and  other  three, 
which,  taken  two  and  two,  have  the  same  ratio ;  if 
the  first  be  greater  than  the  third,  the  fourth  shall  be 
greater  than  the  sixth ;  and  if  equal,  equal  j  and  if 
less,  less. 


OF  EUCLID. 


f45 


Let  A,  B,  C  be  three  magnitudes,  and  D,  E,  F  other  three,  Book  V. 
which,  taken  two  and  two,  have  the  same  ratio,  viz.  as  A  is  to  ^■-y«n^ 
B,  so  is  D  to  E ;  and  as  B  to  C,  so  is  E  to  F.  If 
A  be  greater  than  C,  D  shall  be  greater  than  F; 
and  if  equal,  equal;  and  if  less,  less. 

Because  A  is  greater  than  C,  and  B  is  any 
other  magnitude,  and  that  the  greater  has  to 
the  same  magnitude  a  greater  ratio  than  the  less 
has  to  it*,  therefore  A  has  to  B  a  greater  ratio 
than  C  has  to  B  :  but  as  D  is  to  E,  so  is  A  to  B  ; 
therefore  ^  D  has  to  E  a  greater  ratio  than  C  to 
B ;  and  because  B  is  to  C,  as  E  to  F,  by  inver- 
sion, C  is  to  B,  as  F  is  to  E ;  and  D  was  shown 
to  have  to  E  a  greater  ratio  than  C  to  B ;  there- 
fore D  has  to  E  a  greater  ratio  than  F  toE^  :  but 
the  magnitude  which  has  a  greater  ratio  than 
another  to  the  same  magnitude,  is  the  greater 
of  the  two"*:  D  is  therefore  greater  than  F. 

Secondly,  Let  A  be  equal  to  C ;  D  shall  be  equal  to  F 
cause  A  and  C  are  equal  to  one  an- 


A 
D 


C 

F 


a  8. 5. 


b  13.  5. 


c  Cor. 
13.5. 

d  10.  5, 


be- 


other,  A  is  to  B,  as  C  is  to  B^: 
but  A  is  to  B,  as  D  to  E ;  and  C  is 
to  B,  as  F  to  E ;  wherefore  D  is  to 
E,  as  F  to  Ef;  and  therefore  D  is 
equal  to  F  s. 

Next,  Let  A  be  less  than  C ;  D 
shall  be  less  than  F :  for  C  is  great- 
er than  A,  and,  as  was  shown  in  the 
first  case,  C  is  to  B,  as  F  to  E,  and 
in  like  manner  B  is  to  A,  as  E  to  D  ; 
therefore  F  is  greater  than  D,  by  the 
first  case ;  and  therefore  D  is  less 
than  F.  Therefore,  if  there  be  three, 
&c.    Q.  E.  D. 


A 
D 


B 
E 


C 

F 


A 
D 


B 
E 


t7.S. 

fll.5. 
g9.  5. 


PROP.  XXI.    THEOR. 


IF  there  be   three  magnitudes,  and  other  three,  See  n. 
which  have  the  same  ratio  taken  two  and  two,  but  in 
a  cross  order ;  if  the  first  magnitude  be  greater  than 
the  third,  the  fourth  shall  be  greater  than  the  sixth; 
and  if  equal,  equal;  and  if  less,  less. 

T 


u& 


THE  ELEMENTS 


Book  V, 


h  13.  5. 


c  Cor. 
13.  5. 

d  10.  S. 


f  11.  5. 
S9  5. 


A 
D 


B 
E 


Let  A,  B,  C  be  three  magnitudes,  and  D,  E,  F  other  three, 
'  which  have  the  same  ratio,  taken  two  and  two,  but  in  a  cross 
order,  viz.  as  A  is  to  B,  so  is  E  to  F,  and  as  B 
is  to  C,  so  is  D  to  E.  If  A  be  greater  than  C, 
X)  shall  be  greater  than  F ;  and  if  equal,  equal ; 
and  if  less,  less. 

Because  A  is  greater  than  C,  and  B  is  any 
other  magnitude,  A  has  to  B  a  greater  ratio* 
than  C  has  to  B :  but  as  E  to  F,  so  is  A  to  B ; 
therefore  ^  E  has  to  F  a  greater  ratio  than  C  to 
B :  and  because  B  is  to  C,  as  D  to  E,  by  inver- 
sion, C  is  to  B,  as  E  to  D  :  and  E  was  shown  to 
have  to  F  a  greater  ratio  than  C  to  B ;  there- 
fore E  has  to  F  a  greater  ratio  than  E  to  D  <= ; 
but  the  magnitude  to  which  the  same  has  a 
greater  ratio  than  it  has  to  another,  is  the  lesser 
of  the  two'i ;  F  therefore  is  less  than  D  ;  that  is, 
U  is  greater  than  F. 

Secondly,  Let  A  be  equal  to  C  ;  D  shall  be  equal  to  F.     Be- 
cause A  and  C  are  equal,  A  is  e  to  B,  as  C  is  to  B :  but  A  is 
to  B,  as  E  to  F ;  and  C  is  to  B 
as  E  to  D  ;  wherefore  E  is  to  F 
as  E  to  Df;  and  therefore  D  is 
equal  to  Fs. 

Next,  Let  A  be  less  than  C ; 
D  shall  be  less  than  F :  for  C  is 
greater  than  A,  and,  as  was 
shown,  C  is  to  B,  as  E  to  D, 
and  in  like  manner  B  is  to  A, 
as  F  to  E  ;  therefore  F  is  great- 
er than  D,  by  case  first;  and 
therefore  D  is  less  than  F. 
Therefore,  if  there  be  three. 
Sec.     Q.  E.  D. 


A 
D 


B 
E 


A 
D 


B 

E 


C 

F 


PROP.  XXn.    THEOR. 

SseN.  IF  there  be  any  number  of  magnitudes,  and  as 
many  others,  which,  taken  two  and  two  in  order,  have 
the  same  ratio;  the  first  shall  have  to  the  last  of  the 
first  magnitudes  the  same  ratio  which  the  first  of  the 
others  has  to  the  last.  N.  B.  This  is  usually  died 
by  the  words  *'  ex  aqualiy^''  or  *'  ex  aquo.''^ 


OF  EUCLID. 


14? 


c 

M 


D 
H 


E 
L 


F 

N 


3.4.5-.. 


First,  Let  there  be  three  magnitudes  A,  B,  G,  and  as  many  BookV. 
others  D,  E,  F,  which,  taken  two  and  two,  have  the  same  ratio,  ^m-^^mmj 
that  is,  such  that  A  is  to  B  as  D  to  E ;  and  as  B  is  to  C,  so  is  E 
to  F ;  A  shall  be  to  C,  as  D  to  F. 

Take  of  A  and  D  any  equimultiples  whatever  G  and  H;- 
and  of  B  and  E  any  equimultiples 
whatever  K  and  L ;  and  of  C  and 
F  any  whatever  M  and  N :  then, 
because  A  is  to  B,  as  D  to  E,  and 
that  G,  H  are  equimultiples  of  A, 
D,  and  K,  L  equimultiples  of  B,     A      B 
E ;  as  G  is  to  K,  so  is  »  H  to  L.     G      K 
For  the  same  reason,  K  is  to  M, 
as  L  to  N :  and  because  there  are 
three  magnitudes  G,  K,  M,  and 
other  three  H,  L,  N,  which,  two 
and  two,  have  the  same  ratio ;   if 
G  be  greater  than  M,  H  is  great- 
er than  N ;    and  if  equal,  equal ; 

and  if  less,  less^;    and  G,  Hare  b  20.  §. 

any  equimultiples  whatever  of  A, 
D,  arid  M,  N  are  any  equimul- 
tiples whatever  of  C,  F.      Therefore  c,  as  A  is  to  C,  so  is  D  c  5  de£ 
toF.  ,5. 

Next,  let  there  be  four  magnitudes  A,  B,  C,  D,  and  other 
four  E,  F,  G,  H,  which  two  and  two  have  the 


same  ratio,  viz.  as  A  is  to  B,  so  is  E  to  F,  and     A.  B.  C.  D. 
as  B  to  C,  so  F  to  G ;   and  as  C  to  D,  so  G  to     E.  F.  G.  H. 

H :  A  shall  be  to  D,  as  E  to  H.  1 

Because  A,  B,  C  are  three  magnitudes,  and  E,  F,  G  other 
three,  which,  taken  two  and  two,  have  the  same  ratio ;  by  the 
foregoing  case  A  is  to  C,  as  E  to  G.  But  C  is  to  D,  as  G  is 
to  H ;  wherefore  again,  by  the  first  case,  A  is  to  D,  as  E  to  H : 
and  so  on,  whatever  be  the  number  of  magnitudes.  Therefore, 
if  there  be  any  Humber,  &c.    Q.  E.  D. 


148 

Book  V. 


THE  ELEMENTS 


PROP.  XXIII.    THEOR. 


SeeN. 


IF  there  be  any  number  of  magnitudes,  and  as 
many  others,  which,  taken  two  and  two,  in  a  cross 
order,  have  the  same  ratio,  the  first  shall  have  to  the 
last  of  the  first  magnitudes  the  same  ratio  which  the 
first  of  the  others  has  to  the  last.  N.  B.  This  is  usu- 
ally cited  by  the  words  "  ex  tequali  in  proportione  per- 
*'  turbata  ;^^  or,  "  ex  aquo  perturbate.'*'* 


a  15.  5. 


b  11  5. 


c4.  5. 


d  21.  5. 


First,  Let  there  be  three  magnitudes  A,  B,  C,  and  other  three 
D,  E,  F,  which,  taken  two  and  two,  in  a  cross  order,  have  the 
same  ratio,  that  is,  such  that  A  is  to  B,  as  E  to  F ;  and  as  B  is 
to  C,  so  is  D  to  E :  A  is  to  C,  as  D  to  F. 

Take  of  A.  B,  D  any  equimultiples  whatever  G,  H,  K ;  and 
of  C,  E,  F  any  equimultiples  whatever  L,  M,  N ;  and  because 
G,  H  are  equimultiples  of  A,  B, 
and  that  magnitudes  have  the 
same  ratio  which  their  equimul- 
tiples have»;  as  A  is  to  B,  so  is 
G  to  H.  And,  for  the  same  rea- 
son, as  E  is  to  F,  so  is  M  to  N  : 
but  as  A  is  to  B,  so  is  E  to  F;     A      B      C  D      E     F 

as  therefore  G  is  to  H,  so  is  M  to     G       H      L  KM      N 

N''.  And  because  as  B  is  to  C, 
so  is  D  to  E,  and  that  H,  K  are 
equimultiples  of  B,  D,  and  L,  M 
of  C,  E ;  as  H  is  to  L,  so  is  ^  K 
to  M :  and  it  has  been  shown, 
that  G  is  to  H,  as  M  to  N :  then, 
because  there  are  three  magni- 
tudes G,  H,  L,  and  other  three 
K,  M,  N,  which  have  the  same 
ratio  taken  two  and  two  in  a  cross 
order;    if  G   be   greater  than  L, 

K  is  greater  than  N;  and  if  equal,  equal;  and  if  less,  less**; 
and  G,  K  are  any  equimultiples  whatever  of  A,  D ;  and  L,  N 
any  whatever  of  C,  F;   as,  therefore,  A  is  to  C,  so  is  D  to  F. 


OF  EUCLID. 


140 


A.  B.  C.  D. 

E.  F.  G.  H. 


Next,  Let  there  be  four  magnitudes,  A,  B,  C,  D,  and  other  Book  V. 
four  E,  F,  G,  H,  which  taken  two  and  two 
in  a  cross  order  have  the  same  ratio,  viz. 
A  to  B,  as  G  to  H  ;  B  to  C,  as  F  to  G  ; 
and  C  to  D,  as  E  to  F  :  A  is  to  D  as  E 
toH. 

Because  A,  B,  C  are  three  magnitudes,  and  F,  G,  H  other 
three,  which,  taken  two  and  two  in  a  cross  order,  have  the  same 
ratio  ;  by  the  first  case,  A  is  to  C,  as  F  to  H  :  but  C  is  to  D,  as  E 
is  to  F  ;  wherefore  agani,  by  the  first  case,  A  is  to  D,  as  E  to  H  : 
and  so  on,  whatever  be  the  number  of  magnitudes.  Therefore, 
if  there  be  any  number,  &c.     Q.  E.  D. 


PROP.  XXIV.     THEOR. 

IF  the  first  has  to  the  second  the  same  ratio  which  See  n. 
the  third  has  to  the  fourth ;  and  the  fifth  to  the  se- 
cond the  same  ratio  which  the  sixth  has  to  the  fourth  ; 
the  first  and  fifth  together  shall  have  to  the  second 
the  same  ratio  which  the  third  and  sixth  together 
have  to  the  fourth. 


the  same  ratio  which 
let  BG  the  fifth  have 


B— 


H 


E— 


Let  AB  the  first  have  to  C  the  second 
DE  the  third  has  to  F  the  fourth  ;  and 
to  C  the  second  the  same  ratio  which  EH 
the  sixth  has  to  F  the  fourth  :  AG,  the 
first  and  fifth  together,  shall  have  to  C 
the  second  the  same  ratio  which  DH,  the 
third  and  sixth  together,  has  to  F  the 
fourth. 

Because  BG  is  to  C,  as  EH  to  F  ;  by 
inversion,  C  is  to  BG,  as  F  to  EH  :  and 
because,  as  AB  is  to  C,  so  is  DE  to  F  ;  and 
as  C  to  BG,  so  F  to  EH  ;  ex  aquali  »,  AB 
is  to  BG,  as  DE  to  EH  :  and  because 
these  magnitudes  are  proportionals,  they 
shall  likewise  be  proportionals  when  taken 
jointly  *>  :  as,  therefore,  AG  is  to  G3,  so 

is  DH  to  HE  ;  but  as  GB  to  C,  so  is  HE  to  F.  Tiierefore,  ex 
xquali^^  as  AG  is  to  C,  so  is  DH  to  F.  Wherefore,  if  the  first, 
&c.  Q.  E.  D. 

Cor.  1.  If  the  same  hypothesis  be  made  as  in  the  proposi- 
tbn,  the  excess  of  the  first  and  fifth  shall  be  to  the  second,  as 


A      C      D      F 


a  22.  5. 


bl&5. 


150 


THE   ELEMENTS 


BooRV.  the  excess  of  the  third  and  sixth  to  the  fourth.     The  demonstra- 
^-^v-^ii^  tion  of  this  is  the  same  with  that  of  the  proposition,  if  division 
be  used  instead  of  composition. 

CoR.  2.  The  proposition  holds  true  of  tv/o  ranks  of  magni- 
tudes, whatever  be  their  number,  of  which  each  of  the  first  rank 
has  to  the  second  magnitude  the  same  ratio  that  the  corres- 
ponding one  of  the  second  rank  has  to  a  fourth  magnitude  ;  as 
is  manifest. 


PROP.  XXV.     THEOR. 

IF  four  magnitudes  of  the  same  kind  are  propor- 
tionals, the  greatest  and  least  of  them  together  are 
greater  than  the  other  two  together. 


a  A.  8t 

14.5. 


b  19.  5. 


c  A.  5. 


B 


Let  the  four  magnitudes  AB,  CD,  E,  F  be  proportionals,  viz. 
AB  to  CD,  as  E  to  F  ;  and  let  AB  be  the  greatest  of  them,  and 
consequently  F  the  least ».  AB,  together  with  F,  are  greater 
than  CD,  together  with  E. 

Take  AG  equal  to  E,  and  CH  equal  to  F  :  then,  because  as 
AB  is  to  CD,  so  is  E  to  F,  and  that  AG  is  equal  to  E,  and  CH 
equal  to  F  ;  AB  is  to  CD,  as  AG  to  CH. 
And  because  AB  the  whole  is  to  the 
whole  CD,  as  AG  is  to  CH,  likewise  the 
remainder  GB  shall  be  to  the  remainder 
HD,  as  the  whole  AB  is  to  the  whole  ^ 
CD  :  but  AB  is  greater  than  CD,  there- 
fore «=  GB  is  greater  than  HD  :  and  be- 
cause AG  is  equal  to  E,  and  CH  to  F  ; 
AG  and  F  together  are  equal  to  CH  and 
E  together.  If,  therefore,  to  the  unequal 
magnitudes  GB,   HD,  of  which  GB  is 

the  greater,  there  be  added  equal  magnitudes,  viz.  to  GB  the 
two  AG  and  F,  and  CH  and  E  to  HD ;  AB  and  F  together  are 
greater  than  CD   and   E.     Therefore,  if  four  magnitude?,  8cc. 

Q.  i:.  D. 


G-     D 


H— 


A      C      E      F 


PROP.  F.    THEOR. 


See  N.        RATIOS  which  are  compounded  of  the  same  r^- 
tios,  are  the  same  with  one  another. 


OF  EUCLID. 


151 


A.  B.  C. 
D.  E.  F. 


Let  A  be  to  B  as  D  to  E  ;  and  B  to  C  as  E  to  F  :  the  ratio  which  BookV. 
is  compounded  .of  the  ratio  of  A  to  B, 
and  B  to  C,  which,  by  the  definition  of 
compound  ratio,  is  the  ratio  of  A  to  C,  is 
the  same  with  the  ratio  of  D  to  F,  which, 
by  the  same  definition,  is  compounded  of 
the  ratios  of  D  to  E,  and  E  to  F. 

Because  there  are  three  magnitudes  A,  B,  C,  and  three  others 
D,  E,  F,  which,  taken  two  and  two  in  order,  have  the  same  ra- 
tio ;  ex  xquali,  A  is  to  C  as  D  to  F  3.  a  22.  5. 

Next,  Let  A  be  to  B  as  E  to  F,  and  B  to  C  as  D  to  E  ;  there- 
fore, ex  xqtiali  in  profiortione  fierturbata  ^,  A  is 
to  C  as  D  to  F ;  that  is,  the  ratio  of  A  to  C, 
which  is  compounded  of  the  ratios  of  A  to  B, 
and  B  to  C,  is  the  same  with  the  ratio  of  D  to  F, 
which  is  compounded  of  the  ratios  of  D  to  E, 
and  E  to  F  :  and  in  like  manner  the  proposition  may  be  demons 
strated,  whatever  be  the  nuciber  of  ratios  in  either  case. 


A.  B.  C. 

D.  E.  F. 


b23-5. 


PROP.  G.    THEOR. 

IF  several  ratios  be  the  same  with  several  ratios,  see  ^f. 
each  to  each ;  the  ratio  which  is  compounded  of  ra- 
tios which  are  the  same  with  the  first  ratios,  each  to 
each,  is  the  same  with  the  ratio  compounded  of  ratios 
which  are  the  same  with  the  other  ratios,  each  to  each. 


A.  B.  C.  D. 
E.  F.  G.  H. 


K.  L.  M. 

N.  O.  P. 


Let  A  be  to  B  as  E  to  F  ;  and  C  to  D  as  G  to  H :  and  let 
be  to  B  as  K  to  L  ;  and  C  to  D  as  L  to  M  :  then  the  ratio  of 
to  M,  by  the  definition  of  com- 
pound ratio,  is  compounded  of  the 
ratios  of  K  to  L,  and  L  to  M, 
which  are  the  same  with  the  ra- 
tios of  A  to  B,  and  C  to  D  :  and  as 
E  to  F,  so  let  N  be  to  O ;  and  as  G  to  H,  so  let  O  be  to  P ;  then 
the  ratio  of  N  to  P  is  compounded  of  the  ratios  of  N  to  O,  and 
O  to  P,  which  are  the  same  with  the  ratios  of  E  to  F,  and  G  to 
H:  and  it  is  to  be  shown  that  the  ratio  of  K  to  M  is  the  same 
with  the  ratio  of  N  to  P,  or  that  K  is  to  M  as  N  to  P. 

Because  K  is  to  L  as  (A  to  B,  that  is,  as  E  to  F,  that  is,  as)  N 
t»  O ;  and  as  L  to  M,  so  is  (C  to  D,  and  so  is  G  to  H,  and  so  is) 


152  THE  ELEMENTS 

Book  V.  O  to  P :  ex  isqnali  ^,  K  is  to  M  as  N  to  P.     Therefore,  if  several 
^■i-v^— ;  ratios,  Sec.     Q.  E.  D. 
a  22.  5. 


PROP.  H.    THEOR. 

See  N.  IF  a  rp.tio  compounded  of  several  ratios  be  the  same 
with  a  ratio  compounded  of  any  other  ratios,  and  if 
one  of  the  first  ratios,  or  a  ratio  compounded  of  any 
of  the  first,  be  the  same  witli  one  of  the  last  ratios,  or 
with  the  ratio  compounded  of  any  of  the  last ;  then 
the  ratio  compounded  of  the  remaining  ratios  of  the 
first,  or  the  remaining  ratio  of  the  first,  if  but  one 
remain,  is  the  same  with  the  ratio  compounded  of 
those  remaining  of  the  last,  or  with  the  remaining 
ratio  of  the  last. 

Let  the  first  ratios  be  those  of  A  to  B,  B  to  C,  C  to  D,  D  to  E, 
and  E  to  F ;  and  let  the  other  ratios  be  those  of  G  to  H,  H  to  K, 
K  to  L,  and  L  to  M  :  also,  let  the  ratio  of  A  to  F,  which  is  com- 

a  Defini- pou"fl'-'fl  of*  the  first  ratios,  be  the 

tion  of     same  Avith  the  ratio  of  G  to  M,  which 

com-        is    compounded    of    the    other    ra- 

pounded   jj^g  .  ^^^  besides,  let  the  ratio  of  A 
to  D,  which  is  compounded  of  the 


ratio. 


A.  B.  C.  D.  E.  F. 
G.  H.  K.  L.  M. 


ratios  of  A  to  B,  B  to  C,  C  to  D,  be  the  same  with  the  ratio  of  G 
to  K,  which  is  compounded  of  the  ratios  of  G  to  H,  and  H  to  K  : 
then  the  ratio  compounded  of  the  remaining  first  ratios,  to  wit,  of 
the  ratios  of  D  to  E,  and  E  to  F,  which  compounded  ratio  is  the 
ratio  of  D  to  F,  is  the  same  with  the  ratio  of  K  to  M,  which  is 
compounded  of  the  remaining  ratios  of  K  to  L,  and  L  to  M  of 
the  other  ratios, 
b  B.  5.  Because,  by  the  hypothesis,  A  is  to  D  as  G  to  K,  by  inversion'', 

c  22.  5.  D  is  to  A  as  K  to  G  ;  and  as  A  is  to  F,  so  is  G  to  M  ;  therefore  *=, 
ex  squally  D  is  to  F  as  K  to  iM.  If.  therefore,  a  ratio  which  is,  &c. 
Q.  E.  D. 


OF  EUCLID. 


PROP.  K.    THEOR. 


IF  there  be  any  number  of  ratios,  and  any  number  See  n. 
of  other  ratios  such,  that  the  ratio  compounded  of 
ratios  which  are  the  same  with  the  first  ratios,  each 
to  each,  is  the  same  with  the  ratio  compounded  of 
ratios  which  are  the  same,  each  to  each,  with  the  last 
ratios;  and  if  one  of  the  first  ratios,  or  the  ratio 
which  is  compounded  of  ratios  which  are  the  same 
with  several  of  the  first  ratios,  each  to  each,  be  the 
same  with  one  of  the  last  ratios,  or  with  the  ratio 
compounded  of  ratios  which  are  the  same,  each  to 
each,  with  several  of  the  last  ratios:  then  the  ratio 
compounded  of  ratios  which  are  the  same  with  the 
remaining  ratios  of  the  first,  each  to  each,  or  the  re- 
maining ratio  of  the  firtst,  if  but  one  remain;  is  the 
same  with  the  ratio  compounded  of  ratios  which  are 
the  same  with  those  remaining  of  the  last,  each  to 
each,  or  with  the  remaining  ratio  of  the  last. 

Let  the  ratios  of  A  to  B,  C  to  D,  E  to  F,  be  the  first  ratios ; 
and  the  ratios  of  G  to  H,  K  to  L,  M  to  N,  O  to  P,  Q  to  R, 
be  the  other  ratios  :  and  let  A  be  to  B,  as  S  to  T  ;  and  C  to 
D,  as  T  to  V;  and  E  to  F,  as  V  to  X  :  therefore,  by  the  de- 
finition, of  compound  ratio,  the  ratio  of  S  to  X  is  compounded 


h,  k,I. 
A,  B  ;  C,  D  ;  E,  F. 
G,  H ;  K,  L  ;  M,  N ;  O,  P  ;  Q,  R. 

e,  f,  g.             m,  n,  0,  p. 

S,  T,  V,  X, 
Y,  Z,  a,  b,  c,  d. 

of  the  ratios  uf  S  to  T,  T  to  V,  and  V  to  X,  which  are  the 
same  with  the  ratios  of  A  to  B,  C  to  D,  E  to  F,  each  to  each  ; 
also,  as  G  to  H,  so  let  Y  be  to  Z ;  and  K  to  L,  as  Z  to  a ;  M 
to  N,  as  a  to  b,  O  to  P,  as  b  to  c  ;  and  Q  to  R,  as  c  to  d : 
therefore,  by  the  same  definition,  the  ratio  of  Y  to  d  is  com- 
pounded of  the  ratios  of  Y  to  Z,  Z  to  a,  a  to.  b.  b  to  c,  and 

U 


154 


THE  ELEMENTS,  &c. 


Book  V.  c  to  d,  which  are  the  same,  each  to  each,  with  the  ratios  of  G 
*--v-— ^  to  H,  K  to  L,  M  to  N,  O  to  P,  and  Q  to  R :  therefore,  by  the 
hypothesis,  S  is  to  X,  as  Y  to  d :  also,  let  the  ratio  of  A  to  B, 
that  is,  the  ratio  of  S  to  T,  which  is  one  of  the  first  ratios,  be 
the  same  with  the  ratio  of  e  to  g,  which  is  compounded  of  the 
ratios  of  e  to  f,  and  f  to  g,  which,  by  the  hypothesis,  are  the 
same  with  the  ratios  of  G  to  H,  and  K  to  L,  two  of  the  other 
ratios ;  and  let  the  ratio  of  h  to  1  be  that  which  is  com- 
pounded of  the  ratios  of  h  to  k,  and  k  to  1,  which  are  the  same 
with  the  remaining  first  ratios,  viz.  of  C  to  D,  and  E  to  F ; 
also,  let  the  i-atio  of  m  to  p  be  that  which  is  compounded  of  the 
ratios  of  m  to  n,  n  to  o,  and  o  to  p,  which  are  the  same,  each  to 
each,  Avith  the  remaining  other  ratios,  viz.  of  M  to  N,  O  to  P, 
and  Q  to  R  :  then  the  ratio  of  h  to  1  is  the  same  with  the  ratio  of 
m  to  p,  or  h  is  to  1,  as  m  to  p. 


h,  k,  1. 

A,  B  ;  C,  D  ;  E,  F. 

S,  T,  V,  X. 

G,  H;K,  L;  M,  N ;  O,  P  ;  Q,  R. 

Y,  Z,  a,  b,  c,  d. 

e,  f,  g.             m,  n,  o,  p. 

a  11.  5. 


Because  e  is  to  f,  as  (G  to  H,  that  is,  as)  Y  to  Z ;  and  f  is  to 
g,  as  (K  to  L,  that  is,  as)  Z  to  a ;  therefore,  ex  xquali^  e  is  to  g, 
as  Y  to  a:  and  by  the  hypothesis,  A  is  to  B,  that  is,  S  to  T,  as 
e  to  g  ;  wherefore  S  is  to  T,  as  Y  to  a ;  and,  by  inversion,  T  is  to 
S,  as  a  to  Y ;  and  S  is  to  X,  as  Y  to  d ;  therefore,  ex  (zquali^  T  is 
to  X,  as  a  to  d :  also,  because  h  is  to  k,  as  (C  to  D,  that  is,  as)  T 
to  V  ;  and  k  is  to  1,  as  (E  to  F,  that  is,  as)  V  to  X  ;  therefore,  ex 
aquali^  h  is  to  1,  as  T  to  X :  in  like  manner,  it  may  be  demon- 
strated, that  m  is  to  p,  as  a  to  d  :  and  it  has  been  shown,  that  T 
is  to  X,  as  a  to  d ;  therefore  »  h  is  to  1,  as  m  to  p.     Q.  E.  D. 

The  propositions  G  and  K  are  usually,  for  the  sake  of  brevity, 
expressed  in  the  same  terms  with  propositions  F  and  H :  and 
therefore  it  was  proper  to  show  the  true  meaning  of  them  when 
they  are  so  expressed  ;  especially  since  they  are  very  frequently 
njade  use  of  by  geometers. 


THE 


ELEMENTS  OF  EUCLID. 


BOOK  VI. 


DEFINITIONS. 


I. 

Similar   rectilineal   figures  y\  BookVI. 

are  those  which  have  their  se- 
veral angles  equal,  each  to 
each,  and  the  sides  about  the 
equal  angles  proportionals. 

II. 

"  Reciprocal  figures,  viz.  triangles  and  parallelograms,  are  such  See  IJ: 
"  as  have  their  sides  about  two  of  their  angles  proportionals 
"  in  such  manner,  that  a  side  of  the  first  figure  is  to  a  side 
"  of  the  other,  as  the  remaining  side  of  this  other  is  to  the  re- 
"  maining  side  of  the  first." 

III. 
A  straight  line  is  said  to  be  cut  in  extreme  and  mean  ratio,  when 
the  whole  is  to  the  greater  segment,  as  the  greater  segment  is 
to  the  less. 

IV. 
The  altitude  of  any  figure  is  the  straight  line 
drawn  from  its  vertex  perpendicular  to  the 
base* 


156  THE  ELEMENTS 

Book  VI, 

PROP.  I.    THEOR. 


See  N.        TRIANGLES  and  parallelograms  of  the  same  al- 
titude are  one  to  another  as  their  bases. 


Let  the  triangles  ABC,  ACD,  and  the  parallelograms  EC,  CF, 
have  the  same  altitude,  viz.  the  .perpendicular  drawn  from  the 
point  A  to  BD  :  then,  as  the  base  BC  is  to  the  base  CD,  so  is  the 
triangle  ABC  to  the  triangle  ACD,  and  the  parallelogram  EC  to 
the  parallelogram  CF. 

Produce  BD  both  ways  to  the  points  H,  L,  and  take  any  num- 
ber of  straight  lines  BG,  GH,  each  equal  to  the  base  BC ;  and 
DK,  KL,  any  number  of  them,  each  equal  to  the  base  CD  ;  and 
join  AG,  AH,  AK,  AL  :  then,  because  CB,  BG,  GH  are  all  equal, 

4  38. 1.  the  triangles  AHG,  AGB,  ABC  are  all  equal  a;  therefore,  what- 
ever multiple  the  base  HC  is  of  the  base  BC,  the  same  multiple 
is  the  triangle  AHC  of  the  triangle  ABC:  for  the  same  reason, 
■whatever  multiple  the  base  E      A  F 

LC  is  of  the  base  CD,  the 
same  multiple  is  the  trian- 
gle ALC  of  the  triangle 
ADC :  and  if  the  base  HC 
be  equal  to  the  base  CL, 
the  triangle  AHC  is  also 
equal  to  the  triangle 
ALC  a ;    and  if  the  base   H      G      B     C  D  K         L 

HC  be  greater  than  the  base  CL,  likewise  the  triangle  AHC  is 
greater  than  the  triangle  ALC  ;  and  if  less,  less  :  therefore,  since 
there  are  four  magnitudes,  viz.  the  two  bases  BC,  CD,  and  the 
two  triangles  ABC,  ACD  ;  and  of  the  base  BC  and  the  triangle 
ABC,  the  first  and  third,  any  equimultiples  whatever  have  been 
taken,  viz.  the  base  HC  and  triangle  AHC  ;  and  of  the  base  CD 
and  triangle  ACD,  the  second  and  fourth,  have  been  taken  any 
equimultiples  whatever,  viz.  the  base  CL  and  triangle  ALC  ;  and 
that  it  has  been  shown,  that,  if  the  base  HC  be  greater  than  the 
base  CL,  the  triangle  AHC  is  greater  than  the  triangle  ALC  ; 

hS.delS.  and  if  equal,  equal ;  and  if  less,  less  :  therefoie  •>,  as  the  base  BC 
is  to  the  base  CD,  so  is  the  triangle  ABC  to  the  triangle  ACD. 
And  because  the  parallelogram  CE  is  double  of  the  triangle 


OF  EUCLID.  157 

ABC',   and  the  parallelogram  CF  double  of  the  triangle  ACD,  Book VI. 
and  that  magnitudes  have  the  same  ratio  which  their  equimul-  ^-— v— -^ 
tiples  haved;  as  the  triangle  ABC  is  to  the  triangle  ACD,  so  is  c  41. 1. 
the  parallelogram  EC  to  the  parallelogram  CF  :  and  because  it  d  15.  5. 
has  been  shown,  that,  as  the  base  BC  is  to  the  base  CD,  so  is  the 
triangle  ABC  to  the  triangle  ACD ;  and  as  the  triangle  ABC  to 
the  triangle  ACD,   so  is  the  parallelogram  EC  to  the  parallelo- 
gram CF;  therefore,  as  the  base  BC  is  to  the  base  CD,  so  is^e  11.  5. 
the  parallelogram  EC  to  the  parallelogram  CF.   Wherefore,  tri- 
angles, &c.     Q.  E.  D. 

Cor.  From  this  it  is  plain,  that  triangles  and  parallelograms 
that  have  equal  altitudes,  are  one  to  another  as  tht-ir  bases. 

Let  the  figures  be  placed  so  as  to  have  their  bases  in  the  same 
straight  line  ;  and  having  drawn  perpendiculars  from  the  vertices 
of  the  triangles  to  the  bases,  the  straight  line  which  joins  the 
vertices  is  parallel  to  that  in  which  their  bases  are  ^,  because  the  f  33. 1. 
perpendiculars  are  both  equal  and  parallel  to  one  another:  then, 
if  the  same  construction  be  made  as  in  the  proposition,  the  de- 
monstration will  be  the  same. 


PROP.  n.    THEOR. 

IF  a  straight  line  be  drawn  parallel  to  one  of  the  See  n. 
sides  of  a  triangle,  it  shall  cut  the  other  sides,  or 
those  produced,  proportionally :  and  if  the  sides,  or 
the  sides  produced,  be  cut  proportionally,  the  straight 
line  which  joins  the  points  of  section  shall  be  paral- 
lel to  the  remaining  side  of  the  triangle. 

Let  DE  be  drawn  parallel  to  BC,  one  of  the  sides  of  the  trian- 
gle ABC  :  BD  is  to  DA,  as  CE  to  EA. 

Join  BE,  CD;  then  the  triangle  BDE  is  equal  to  the  triangle 
CDEa,  because  they  are  on  the  same  base  DE,  and  between  the  a  37. 1. 
same  parallels  DE,  BC:  ADE  is   another  triangle,  and  equal 
magnitudes  have  to  the  same  the  same  ratio^;  therefore,  as  the  b  7.  5. 
triangle  BDE  to  the  triangle  ADE,  so  is  the  triangle  CDE  to 
the  triangle   ADE;  but   as  the   triangle  BDE   to  the   triangle 
ADE,  so  isc  BD  to  DA,  because  haviug  the  same  altitude,  viz.  c  1.  6. 
the  perpendicular  drawn  from  the  point  E  to  AB,  they  are  to  one 
another  as  their  bases;  and  for  the  same  reason,  as  ihe  triangle 


158 


THE  ELEMENTS 


BookVI.CDE  to  the  triangle  ADE,   so  is  E  to  EA.    Therefore,  as  BD 

»— V— '  to  DA,  so  is  CE  to  EAd. 

d  11.5.        Next,  Let  the  sides  AB,  AC  of  the  triangle  ABC,  or  these 


el.  6. 


f9.  5. 


g  39. 1. 


produced,  be  cut  proportionally  in  the  points  D,  E,  that  is,  so 
that  BD  be  to  DA,  as  CE  to  EA,  and  join  DE  ;  DE  is  parallel 
to  BC. 

The  same  construction  being  made,  because  as  BD  to  DA, 
so  is  CE  to  EA ;  and  as  BD  to  DA,  so  is  the  triangle  BDE  to 
the  triangle  ADE*;  and  as  CE  to  EA,  so  is, the  triangle  CDE 
to  the  triangle  ADE ;  therefore  the  triangle  BDE  is  to  the 
triangle  ADE,  as  the  triangle  CDE  to  the  triangle  ADE ; 
that  is,  the  triangles  BDE,  CDE  have  the  same  ratio  to  the  tri- 
angle ADE  ;  and  therefore  f  the  triangle  BDE  is  equal  to  the  tri- 
angle CDE :  and  they  are  on  the  same  base  DE ;  but  equal  tri- 
angles on  the  same  base  are  between  the  same  parallels  s  ;  there- 
fore DE  is  parallel  to  BC.  Wherefore,  if  a  straight  line.  Sec. 
Q.  E.  D. 

PROP.  in.     THEOR. 


See  N  IF  the  angle  of  a  triangle  be  divided  into  two  equal 
angles,  by  a  straight  line  which  also  cuts  the  base;  the 
segments  of  the  base  shall  have  the  same  ratio  which 
the  other  sides  of  the  triangle  have  to  one  another : 
and  if  the  segments  of  the  base  have  the  same  ratio 
which  the  other  sides  of  the  triangle  have  to  one  ano- 
ther, the  straight  line  drawn  from  the  vertex  to  the 
point  of  section  divides  the  vertical  angle  into  two 
equal  angles. 

Let  the  angle  BAC  of  any  triangle  ABC  be  divided  into  two 
equal  angles  by  the  straight  line  AD  :  BD  is  to  DC,  as  B  A  to  \C. 


OF  EUCLID.  15^ 

Through  the  point  C  draw  CE  parallel  *  to  DA,  and  let  BA  Book VI- 
produced  meet  CE  in  E.     Because  the  straight  line  AC  meets  ^— v— ' 
the  parallels  AD,  EC,  the  angle  ACE  is  equal  to  the  alternate  a  31. 1. 
angle  CAD  ^  :  but  CAD,  by  the  hypothesis,  is  equal  to  the  angle  b  29. 1- 
BAD  ;  wherefore  BAD  is  equal  to  the  angle  ACE.     Again,  be- 
cause  the   straight   line    BAE  E 
meets  the  parallels  AD,  EC,  the 
outward  angle  BAD  is  equal  to                                 ^ 
the  inward  and  opposite  angle 
AEC :     but    the    angle   ACE 
has  been  proved    equal  to  the 
angle     BAD ;     therefore    also 
ACE  is  equal  to  the  angle  AEC, 
and  consequently  the  side  AE 

is  equal  to  the  side  <=■  AC ;  and     •"  xy  v.  c  6. 1. 

because  AD  is  drawn  parallel  to  one  of  the  sides  of  the  triangle 
BCE,  viz.  to  EC,  BD  is  to  DC  as  BA  to  AEd  ;  but  AE  is  equal  d  2.  6. 
to  AC  ;  therefore,  as  BD  to  DC,  so  is  BA  to  AC  «.  e  7.  5. 

Let  now  BD  be  to  DC  as  BA  to  AC,  and  join  AD  ;  the  angle 
BAC  is  divided  into  two  equal  angles  by  the  straight  line  AD. 

The  same  construction  being  made  ;  because,  as  BD  to  DC,  so 
is  BA  to  AC  ;  and  as  BD  to  DC,  so  is  BA  to  AE  ^,  because  AD 
is  parallel  to  EC  ;  therefore  BA  is  to  AC  as  BA  to  AE  f :  conse-f  11.3; 
quently  AC  is  equal  to  AE  s,  and  the  angle  AEC  is  therefore  g  9.  5. 
equal  to  the  angle  ACE  ^  :  but  the  angle  AEC  is  equal  to  the  out-  h  5. 1, 
ward  and  opposite  angle  BAD :  and  the  angle  ACE  is  equal  to 
the  alternate  angle  CAD'^:  wherefore  also  the  angle  BAD  is 
equal  to  the  angle  CAD  :  therefore  the  angle  BAC  is  cut  into 
two  equal  angles  by  the  straight  line  AD.     Therefore,  if  the  an- 
gle, 8cc.     Q.  E.  D. 


THE  ELEMENTS 


PROP.  A.    THEOR. 


IF  the  outward  angle  of  a  triangle,  made  by  produ- 
cing one  of  its  sides,  be  divided  into  two  equal  angles, 
by  a  straight  line  which  also  cuts  the  base  produced; 
the  segments  between  the  dividing  line  and  the  ex- 
tremities of  the  base  have  the  same  ratio  which  the 
other  sides  of  the  triangle  have  to  one  another  :  and 
if  the  segments  of  the  base  produced,  have  the  same 
ratio  which  the  other  sides  of  the  triangle  have,  the 
straight  line  drawn  from  the  vertex  to  the  point  of 
section  divides  the  outward  angle  of  the  triangle  into 
two  equal  angles. 


Let  the  outward  angle  CAE  of  any  triangle  ABC  be  divided 
into  two  equal  angles  by  the  straight  line  AD*  which  meets  the 
base  produced  in  I)  :  BD  is  to  DC  as  BA  to  AC. 

a  31. 1.  Throu,i^h  C  draw  CF  parallel  to  AD  » :  and  because  the  straight 
line  AC  meets  the  parallels  AD,  FC,  the  angle  ACF  is  equal  to 

b  29. 1.    the  alternate  angle  CAD**:    but  CAD   is  equal   to  the    angle 

c  Hyp.  DAE  c  ;  therefore  also  DAE  is  equal  to  the  angle  ACF.  Again, 
because  the  straight  line  FAE  meets  the  parallels  AD,  FC,  the 
outward  angle  DAE  is  equal  ~ 

to  the  inward  and  opposite 
angle  CFA:  but  the  angle 
ACF  has  been  proved  equal 
to  the  angle  DAE;  therefore 
also  the  angle  ACF  is  equal 
to  the  angle  CP'A,  and  conse- 
quently the  side  AF  is  equal 

d  6.  1.      to  the  side  AC  ^  :  and  because 

AD  is  parallel  to  FC,  a  side  of  the  triangle  BCF,  BD  is  to  DC  as 

e  2.  6.  BA  to  AF^  ;  but  AF  is  equal  to  AC  ;  as  therefore  BD  is  to  DC, 
so  is  BA  to  AC. 

Let  now  BD  be  to  DC  as  BA  to  AC,  and  join  AD  ;  the  angle 
CAD  is  equal  to  the  angle  DAE. 

The  same  construction  being   made,   because   BD  is  to  DC 

fU.  5.     as  BA  to  AC;   and  that   BD   is   also   to   DC   as   BA  to  AF  f ; 

g  9.  Tf.      tiierefore  BA  is  to  AC  as  BA  to  AF  e  ;  wherefore  AC  is  equal 

hS.l.      to   AFh,  and  the  angle  AFC  equal  *»  to  the  angle  ACF:  but 


OF  EUCLID.  IS  I 

the  angle  AFC  is  equal  to  the  outward  angle  EAD,  and  the  Book  VI. 
angle  ACF  to  the  alternate  angle  CAD  ;  therefore  also  EAD  is  *— y^ 
equal    to   the   angle    CAD.     Wherefore,   if  the   outward,  &c. 
Q.  E.  D. 


PROP.  IV.    THEOR. 

THE  sides  about  the  equal  angles  of  equiangular 
triangles  are  proportionals ;  and  those  which  are  op- 
posite to  the  equal  angles  are  homologous  sides,  that 
is,  are  the  antecedents  or  consequents  of  the  ratios. 

Let  ABC,  DCE  be  equiangular  triangles,  having  the  angle 
ABC  equal  to  the  angle  DCE,  and  the  angle  ACB  to  the  angle 
DEC,  and  consequently »  the  angle  BAC  equal  to  the  angle  a  32. 1. 
CDE.  The  sides  about  the  equal  angles  of  the  triangles  ABC, 
DCE  are  proportionals  ;  and  those  are  the  homologous  sides 
which  are  opposite  to  the  equal  angles. 

Let  the  triangle  DCE  be  placed,  so  that  its  side  CE  may  be 
contiguous  to  BC,  and  in  the  same  straight  line  with  it :  and 
because  the  angles  ABC,  ACB  are  together  less  than  two  right 
angles  b,  ABC,  and  DEC,  which  is     F  Jn^  b  17.1. 

equal  to  ACB,  are  also  less  than  two 
right  angles  ;  wherefore  BA,  ED 
produced  shall  meet  <= ;  let  them  be 
produced  and  meet  in  the  point  F  : 
and  because  the  angle  ABC  is  equal 

to  the  angle  DCE,  BF  is  parallel  ^  \  ^\.,^    \      \^      d28. 1, 

to  CD.  Again,  because  the  angle 
ACB  is  equal  to  the  angle  DEC, 
AC  is   parallel  to  FE  d :  therefore 

FACD   is  a  parallelogram  ;  and   consequently  AF  is  equal  to 
CD,  and  AC  to  FD  «  :  and  because  AC  is  parallel  to  FE,  one  of  e  34. 1. 
the  sides  of  the  triangle  FBE,  BA  is  to  AF,  as  BC  to  CE  f :  but  f  2.  6. 
AF  is  equal  to  CD  ;  therefore  s,  as  BA  to  CD,  so  is  BC  to  CE  ;  g  7.5. 
and  alternately,  as  AB  to  BC,  so  is  DC  to  CE*  :  again,  because 
CD  is  parallel  to  BF,  as  BC  to  CE,  so  is  FD  to  DE^ :  but  FD  is 
equal  to  AC  ;  therefore,  as  BC  to  CE,  so  is  AC  to  DE :  and   al- 
ternately, as  BC  to  CA,  so  CE  to  ED  :  therefore,  because  it  has 
been  proved  that  AB  is  to  BC,  as  DC  to  CE,  and  as  BC  to  CA,  so 
CE  to  ED,  ex  aquuli\  B  A  i.s  to  AC,  as  CD  to  DE.     Therefore,  h  22.  5.. 
the  sides,  &:c.     Q.  E.  D. 

X 


162  THE  ELEMENTS 

Book  VI. 

PROP.  V.    THEOR. 


IF  the  sides  of  two  triangles,  about  each  of  theii* 
angles,  be  proportionals,  the  triangles  shall  be  equi- 
angular, and  have  their  equal  angles  opposite  to  the 
homologous  sides. 


Let  the  triangles  ABC,  DEF  have  their  sides  proportionals, 
so  that  A  B  is  to  BC,  as  DE  ^o  EF;  and  BC  to  CA,  as  EF  to 
FD ;  and  consequently,  ex  (equali,  BA  to  AC,  as  ED  to  DF  ; 
the  triangle  ABC  is  equiangular  to  the  triangle  DEF,  and  their 
equal  angles  are  opposite  to  the  homologous  sides,  viz.  the  angle 
ABC  equal  to  the  angle  DEF,  and  BCA  to  EFD,  and  also  BAC 
to  EDF. 
a  23. 1.  At  the  points  E,  F,  in  the  straight  line  EF,  make  »  the  angle 
FEG  equal  to  the  angle  ABC,  and  the  angle  EFG  equal  to 
BCA  ;  wherefore  the  remain-  ~ 

ing  angle  BAC  is  equal  to  the 
b32. 1.  remaining  angle  EGF^,  and 
the  triangle  ABC  is  therefore 
equiangular  to  the  triangle 
GEF  ;  ana  consequently  they 
have  their  sides  opposite  to 
the  equal  angles  proportion- 
ed 6.      alsc.     Wherefore,  as   AB    to 

BC,  so  is  GE  to  EF  ;  but  as  AB  to  BC,  so  is  DE  to  EF ;  there- 
d  11.  5.    fore  as  DE  to  EF,  so^  GE  to  EF  :  therefore  DE   and  GE  have 
e  9.  5.      the  same  ratio  to  EF,  and  consequently  are  equal  e  :  for  the  same 
reason,  DF  is  equal  to  FG :  and  because  in  the  triangles  DEF, 
GEF,  DE  is  equal  to  EG,  and  EF  common,  the  two  sides  DE, 
EF  are  equal  to  the  two  GE,  EF,  and  the  base  DF  is  equal  to  the 
f  8. 1.       bass  GF  ;  therefore  the  angle  DEF  is  equal f  to  the  angle  GEF, 
and  the  other  angles  to  the  other  angles  which  are  subtended  by 
S  *•  ^-      the  equal  sides  e.    Wherefore  the  angle  DFE  is  equal  to  the  an- 
gle GFE,and  FDF  to  EGF  :  and  because  the  angle  DEF  is  equal 
to  the  angle  GEF,  and  GEF  to  the  angle  ABC  ;  therefore  the 
angle  ABC  is  equal  to  the  angle  DEF  :  for  the  same  reason,  the 
angle  ACB  is  equal  to  the  angle  DFE,  and  the  angle  at  A  to  the 
angle  at  D.     Thert^fore  the  triangle  ABC  is  equiangular  to  the 
triangle  DEF.    Wherefore,  if  the  sides,  &c.     Q.  E.  D. 


OF  EUCLID. 


PROP.  VI.     THEOR. 


IF  two  triangles  have  one  angle  of  the  one  equal 
to  one  angle  of  the  other,  and  the  sides  about  the  equal 
angles  proportionals,  the  triangles  shall  be  equiangu- 
lar, and  shall  have  those  angles  equal  which  are  oppo- 
site  to  the  homologous  sides. 

Let  the  triangles  ABC,  DEF  have  the  angle  BAC  in  the  one 
equal  to  the  angle  EDF  in  the  other,  and  the  sides  about  those 
angles  proportionals  ;  that  is,  BA  to  AC,  as  ED  to  DF ;  the 
triangles  ABC,  DEF  are  equiangular,  and  have  the  angle  ABC 
equal  to  the  angle  DEF,  and  ACB  to  DFE. 

At  the  points  D,  F,  in  the  straight  line  DF,  make  »  the  angle  a  23. 1. 
FDG  equal  to  either  of  the  angles  BAC,  EDF  ;  and  the  angle 
DFG  equal  to  the  angle  ACB  ;  A 
wherefore    the    remaining   an-    \  j) 

gle    at  B  is  equal  to  the    re-    I  \  /V— — — _'  G 

maining  one  at  G  •>,  and  con-  |      \  /  \^      '     1     b32. 1. 

scquently  the  triangle  ABC  is 
equiangular  to  the  triangle 
DGF  ;  and  therefore  as  BA  to 

AC,  so  is«   GD  to  DF:   hnt,/  \        I N^        c4.6. 

by   the   hypothesis,    as  BA  to  B  C        E  F 

AC,  so  is  ED  to  DF ;  as  therefore  ED  to  DF,  so  is  4  GD  to  DF  ;  «1 11-  ?• 
wherefore  ED  is  equal  e  to  DG  ;   and  DF  is  common  to  the  two  e  9.  5. 
triangles  EDF,  GDF  :  therefore  the  two  sides  ED,  DF  are  equal 
to  the  two  sides  GD,  DF ;  and  the  angle  EDF  is  equal  to  the 
angle  GDF  ;  wherefore  the  base  EF  is  equal  to  the  base  FGf,  **-^- 
and  the  triangle  EDF  to  the  triangle  GDF,  and  the  remaining 
angles  to  the  remaining  angles,  each  to  each,  which  are  subtend- 
ed by  the  equal  sides :  therefore  the  angle  DFG  is  equal  to  the 
angle  DFE,  and  the  angle  at  G  to  the  angle  at  E  :  but  the  angle 
DFG  is  equal  to  the  angle  ACB ;  therefore  the   angle   ACB  is 
equal  to  the  angle  DFE :  and  the  angle  BAC  is  equal  to  the  an- 
gle EDF  g  ;  wherefore  also  the  remaining  angle  at  B  is  equal  to  S  Hyp- 
the  remaining  angle  at  E.     Therefore  the  triangle  ABC  is  equi- 
angular to  the  triangle  DEF.     Wherefore,  if  two  triangles,  8cq. 
Q.  E.  D. 


J.64 
Book  VI. 


THE  ELEMENTS 


PRO?.  VII.    THEOR. 


See  N. 


a  23. 1. 


b52.1. 

c4.  6. 

d  11.  5. 
e9.  5. 

f5. 1. 
g  13. 1. 


IF  two  triangles  have  one  angle  of  the  one  equal 
to  one  angle  of  the  other,  and  the  sides  about  two 
other  angles  proportionals,  then,  if  each  of  the  re- 
maining angles  be  either  less,  or  not  less,  than  a  right 
angle;  or  if  one  of  them  be  a  right  angle:  the  trian- 
gles shall  be  equiangular,  and  have  those  angles  equal 
about  which  the  sides  are  proportionals. 

Let  the  two  triangles  ABC,  DEF  have  one  angle  in  the  one 
equal  to  one  angle  in  the  other,  viz.  the  angle  BAC  to  the  angle 
EDF,  and  the  sides  about  two  other  angles  ABC,  DEF  propor- 
tionals, so  that  AB  is  to  BC,  as  DE  to  EF  ;  and,  in  the  first 
case,  let  each  of  tlie  remaining  angles  at  C,  F  be  less  than  a  right 
angle.  The  triangle  ABC  is  equiangular  to  the  triangle  DEF, 
viz.  the  angle  ABC  is  equal  to  the  angle  DEF,  and  the  remain- 
ing angle  at  C  to  the  remaining  angle  at  F. 

For,  if  the  angles  ABC,  DEF  be  not  equal,  one  of  them  is 
greater  than  the  other:  let  ABC  be  the  greater,  and  at  the 
point  B,  in  the  straight  line 
AB,  make  the  angle  ABG 
equal  to  the  angle  ^  DEF  : 
and  because  the  angle  at  A 
is  equal  to  the  angle  at  D, 
and  the  angle  ABG  to  the 
angle  DEF  ;  the  remaining 
angle  AGB  is  equal  ^  to  the 

remaining  angle  DFE :  therefore  the  triangle  ABG  is  equi- 
angular to  the  triangle  DEF  ;  wherefore  <=  as  AB  is  to  BG,  so  is 
DE  to  EF  ;  but  as  DE  to  EF,  so,  by  hypothesis,  is  AB  to  BC  ; 
therefore  as  AB  to  BC,  so  is  AB  to  BG  d ;  and  because  AB 
has  the  same  ratio  to  each  of  the  lines  BC,  BG  ;  BC  is  equal  * 
to  BG,  and  therefore  tlie  angle  BGC  is  equal  to  the  angle 
BCGf:  but  the  angle  BCG  is,  by  hypothesis,  less  than  a  right 
angle  ;  therefore  also  the  angle  BGC  is  less  than  a  right  angle, 
and  the  adjacent  angle  AGB  must  be  greater  than  a  right  angled. 
But  it  was  proved  that  the  angle  AGB  is  equal  to  the  angle  at  F  ; 
therefore  the  angle  at  F  is  greater  than  a  right  angle  :  but  by  the 
hypothesis,  it  is  less  than  a  right  angle ;  which  is  absurd.   There- 


OF  EUCLID. 


165 


fore  the  angles  ABC,  DEF  are  not  unequal,  that  is,  they  are  Book  VI. 
equal :  and  the  angle  at  A  is  equal  to  the  angle  at  D ;  where-  ^-^v— ^ 
fore  the  remaining  angle  at  C  is  equal  to  the  remaining  angle  at 
F  :  therefore  the  triangle  ABC  is  equiangular  to  the  triangle  DEF. 

Next,  Let  each  of  the  angles  at  C,  F,  be  not  less  than  a  right 
angle :  the  triangle  ABC  is  also  in  this  case  equiangular  to  the 
triangle  DEF. 

The  same  construction  being  A 

made,  it  may  be  proved  in  like  a.  D 

manner  that  BC  is  equal  to  BG, 
and  the  angle  at  C  equal  to  the 
angle  BGC :  but  the  angle  at  C 
is  not  less  than  a  right  angle ; 
therefore  the  angle  BGC  is  not 
less  than  a  right  angle :   wherefore  two  angles  of  the  triangle 
BGC  are  together  not  less  than  two  righl  angles,  which  is  im- 
possible*'; and  therefore  the  triangle  ABC  may  be  proved  to  behl71- 
equiangular  to  the  triangle  DEF,  as  in  the  first  case. 

Lastly,  Let  one  of  the  angles  at  C,  F,  viz.  the  angle  at  C,  be  a 
right  angle ;  in  this  case  likewise  the  triangle  ABC  is  equiangu- 
lar to  the  triangle  DEF. 

For,  if  they  be  not  equiangu-  A 

lar,  make,  at  the  point  B  of  the 
straight  line  AB,  the  angle  ABG 
equal  to  the  angle  DEF ;  then  it 
may  be  proved,  as  in  the  first 
case,  that  BG  is  equal  to  BC: 
but  the  angle  ECG  is  a  right 
angle,  therefore'  the  angle  BGC 
is  also  a  right  angle ;  whence 
two  of  the  angles  of  the  trian- 
gle BGC  are  together  not  less 
than  two  right  angles,  which  is 
)mpossiblel»:  therefore  the  tri-  B 
angle  ABC  is  equiangular  to  the 
triangle  DEF.  Wherefore,  if 
two  triangles,  &c.     Q.  E.  D. 


i  5. 1. 


THE   ELEMENTS 


PROP.  VIII.    THEOR. 


SeeN 


IN  a  right  angled  triangle,  if  a  perpendicular  be- 
drawn  from  the  right  angle  to  the  base,  the  triangles 
on  each  side  of  it  are  similar  to  the  whole  triangle, 
and  to  one  another. 


a  32.1 


b4.  6. 


Let  ABC  be  a  right  angled  triangle,  having  the  right  angle 
BAC  ;  and  from  the  point  A  let  AH  be  drawn  perpendicular  to 
the  base  BC :  the  triangles  ABD,  ADC  are  similar  to  the  whole 
triangle  ABC,  and  to  one  another. 

Because  the  angle  BAC  is  equal  to  the  angle  ADB,  each  of 
them  being  a  right  angle,  and  that  the  angle  at  B  is  common 
to  the  two  triangles  ABC,  ABD;  , 

the  remaining  angle  ACB  is  equal 
to  the  remaining  angle  BAD*: 
therefore  the  triangle  ABC  is  equi- 
angular to  the  triangle  ABD,  and 
the  sides  about  their  equal  angles 
are  proportionals'';  wherefore  the 
cl.def.6.  trjjtngles  are  similar^:  in  the  like 
manner  it  may  be  demonstrated, 
that  the  triangle  ADC  is  equiangular  and  similar  to' the  triangle 
ABC;  and  the  triangles  ABD,  ADC,  being  both  equiangular 
and  similar  to  ABC,  are  equiangular  and  similar  to  each  other. 
Therefore,  in  a  right  angled.  Sec.     Q.  E.  D. 

CoH.  From  this  it  is  manifest,  that  the  perpendicular  drawn 
from  the  right  angle  of  a  right  angled  triangle  to  the  base,  i,s  a 
mean  proportional  between  the  segments  of  the  base :  and  also 
that  each  of  the  sides  is  a  mean  proportional  between  the  base, 
and  Its  segment  adjacent  to  that  side:  because  in  the  triangles 
BDA,  ADC,  BD  is  to  DA,  as  DA  to  DC  •» ;  and  in  the  triangles 
ABC,  DBA,  BC  is  to  BA,  as  BA  to  BD»>;  and  in  the  triangle?. 
ABC,  ACDj  BC  is  to  CA^  as  CA  to  CDi>. 


OF  EUCLID. 


PROP.  IX.    PROB. 


FROM  a  given  straight  line  to  cut  off  any  part  See  n. 
required. 


Let  AB  be  the  given  straight  line  ;  it  is  required  to  cut  off  any 
part  from  it. 

From  the  point  A  draw  a  straight  line  AC  making  any  angle 
with  AB ;  and  in  AC  take  any  point  D,  and  take  AC  the  same 
multiple  of  AD,  that  AB  is  of  the  part  which 
is  to  be  cut  off  from  it ;  join  BC,  and  draw  A 

DE  parallel  to  It :  then  AE  is  the  part  requir- 
ed to  be  cut  off. 

Because  ED  is  parallel  to  one  of  the  sides 
of  the  triangle  ABC,  viz.  to  BC,  as  CD  is  to 
DA,  so  is  *  BE  to  EA  ;  and,  by  composition  t, 
CA  is  to  AD  as  BA  to  AE :  but  CA  is  a  mul- 
tiple of  AD  ;  therefore  c  BA  is  the  same  mul- 
tiple of  AE  :  whatever  part  therefore  AD  is 
of  AC,  AE  is  the  same  part  of  AB  :  where- 
fore, from  the  straight  line  AB  the  part  re- 
quired is  cut  off.     Which  was  to  be  done. 


PROP.  X.     PROB. 


TO  divide  a  given  straight  line  similarly  to  a  given 
divided  straight  line,  that  is,  into  parts  that  shall  have 
the  same  ratios  to  one  another  which  the  parts  of  the 
divided  given  straight  line  have. 

Let  AB  be  the  straight  line  given  to  be  divided,  and  AC  the 
divided  line ;  it  is  required  to  divide  AB  similarly  to  AC. 

Let  AC  be  divided  in  the  points  D,  E ;  and  let  AB,  AC  be 
placed  so  as  to  contain  any  angle,  and  join  BC,  and  through  the 
points  D,   E  draw  »  DP,  EG  parallels  to  it ;    and  through   D  a  31.  1. 
draw  DHK  parallel  to  AB  :  therefore  each  of  the  figures  FH, 
IIB,  is  a  parallelogram;  wherefore  DH  is  equal  ^  to  FO,  andbSi.  1 


168 


THE  ELEMENTS 


Book  VI.  HK  to  GB:  and  because  HE  is  paral- 

*— ■V-— ^  lei  to  KC,  one  of  the  sides  of  the  trian- 

c  2.  6.  gle  DKC,  as  CE  to  ED,  so  is  c  KH  to 
HD  :  but  KH  is  equal  to  BG,  and  HD 
to  GF ;  therefore  as  CE  to  ED,  so  is 
BG  to  GF  :  again,  because  FD  is  pa- 
rallel to  EG,  one  of  the  sides  of  the  tri- 
angle AGE,  as  ED  to  DA,  so  is  GF  to 
FA  ;  but  it  has  been  proved  that  CE  is 
to  ED  as  BG  to  GF;  and  as  ED  to  DA, 
so  GF  to  FA  :  therefore  the  given  straight  line  AB  is  divided  si- 
milarly to  AC.     Which  was  to  be  done. 


PROP.  XI.     PROB. 


TO  find  a  third  proportional  to  two  given  straight 
lines* 


a31.  1. 


b  2.  6. 


Let  AB,  AC  be  the  two  given  straight  lines,  and  let  them  be 
placed  so  as  to  contain  any  angle  ;  it  is  requir- 
ed to  find  a  third  proportional  to  AB,  AC. 

Produce  AB,  AC  to  the  points  D,  E  ;  and 
make  BD  equal  to  AC  ;  and  having  joined  BC, 
through  D  draw  DE  parallel  to  it». 

Because  BC  is  parallel  to  DE,  a  side  of  the 
triangle  ADE,  AB  is  ^  to  BD  as  AC  to  CE  : 
but  BD  is  equal  to  AC;  as  therefore  AB  to  AC, 
so  is  AC  to  CE.  Wherefore,  to  the  two  given 
straight  lines  AB,  AC  a  third  proportional  CE 
is  found.     Which  was  to  be  done. 


PROP.  Xn.    PROB. 


TO  find  a  fourth  proportional  to  three  given  straight 
lines. 


Let  A,  B,  C  be  the  three  given  straight  lines ;  it  is  required  to 
find  a  fourth  j)roportional  to  A,  B,  C 


OF  EUCLID. 


169 


Take  two  straight  lines  DE,  DF,  containing  any  angle  EDF  ;  Book  VI 
and  upon  these  make  DG  equal 
to  A,  GE  equal  to  B,  and  DH 
equal  to  C  ;  and  havint^  joined 
GH,  draw  EF  parallel »  to  it 
through  the  point  E  :  and  be- 
cause GH  is  parallel  to  EF,  one 
of  the  sides  of  the  triangle 
DEF,  DG  is  to  GE,  as  DH  to 
HFb;  but  DG  is  equal  to  A, 
GE  to  B,  and  DH  to  C  ;  there- 
fore, as  A  is  to  B,  so  is  C  to  HF. 

Wherefore  to  the  three   given  straight  lines  A,  B,  C  a  fourth 
proportional  HF  is  found.     Which  was  to  be  done. 


b%G, 


PROP.  XIII.    PROB. 


TO  find  a  mean  proportional  between  two  given 
straight  lines. 

Let  AB,  BC  be  the  two  given  straight  lines  ;  it  is  required  to 
find  a  mean  proportional  between  them. 

Place  AB,  BC  in  a  straight  line,  and  upon  AC  descrih'^  the 
semicircle  ADC,  and  from  the 
point  B  draw  *  BD  at  right  an- 
gles to  AC,  and  join  AD,  DC. 

Because  the  angle  ADC  in  a 
semicircle  is  a  right  angle  *>,  and 
because  in  the  right  angled  tri- 
angle ADC,  DB  is  drawn  from 
the  right  angle  perpendicular  to 
the  base,  DB  is  a  mean  propor- 
tional between  AB,  BC,  the  segments  of  the  base  <= :  therefore  be-  c  Cor.  8, 
tween  the  two  given  straight  lines  AB,  BC  a  mean  proportional  ^ 
DB  is  found.    Which  was  to  be  done. 


a  11. 1. 


b  31.  3, 


THE  ELEMENTS 


PROP.  XIV.    THEOR. 


EQUAL  parallelograms  which  have  one  angle  of 
the  one  equal  to  one  angle  of  the  other,  have  their 
sides  about  the  equal  angles  reciprocally  proportion- 
al :  and  parallelograms  that  have  one  angle  of  the  one 
equal  to  one  angle  of  the  other,  and  their  sides  about 
the  equal  angles  reciprocally  proportional,  are  equal 
to  one  another. 


Let  AB,  BC  be  equal  parallelosframs,  which  have  the  angles 
at  B  equal,  and  lee  the  sides   DB,  BE  be  placed  in  the  same 

a  14.  1*  straight  line  ;  wherefore  also  FB,  BG  are  in  one  straight  line  »  : 
the  sides  of  the  parallelograms  AB,  BC  about  the  equal  angles, 
are  reciprocally  proportional ;  that  is,  DB  is  to  BE,  as  GB  to  BF. 
Complete  the  parallelogram  FE  ;  and  because  the  parallelo- 
gram AB  is  equal  to  BC,  and  that  A  F 
FE  is  another  parallelogram,   AB 

^7.5.  is  to  FE,  as  BC  to  FE^:  but  as 
AB  to  FE,  so  is  the  base  DB  to 

c  1.  6.  BE  c  ;  and,  as  BC  to  FE,  so  is  the 
base  GB  to  BF  ;  therefore,  as  DB 

•i  11-  ^'  to  BE,  so  is  GB  to  BF  d.  Where- 
fore the  sides  of  the  parallelograms 
AB,  BC  about  their  equal  angles 
are  reciprocally  proportional. 

But,  let  the  sides  about  the  equal  angles  be  reciprocally  pro- 
portional, viz.  as  DB  to  BE,  so  GB  to  BF  ;  the  parallelogram  AB 
is  equal  to  the  parallelogram  BC. 

Because,  as  DBto  BE,  so  is  GB  to  BF  ;  and  as  DB  to  BE,  so  is 
the  parallelogram  AB  to  the  parallelogram  FE  ;  and  as  GB  to 
BF,  so  is  the  parallelogram  BC  to  the  parallelogram  FE  ;  there- 
fore as  AB  to  FE,  so  BC  to  FE  ^ :  wherefore  the  parallelogram 

*  ^-  ■^'      AB  is  equal «  to  the  parallelogram  BC.     Therefore,  equal  paral. 
lelogramsj  Sec.    Q.  E.  D. 


OF  EUCLID. 


PROP.  XV.    THEOR. 


EQUAL  triangles  which  have  one  angle  of  the  one 
equal  to  one  angle  of  the  other,  have  their  sides  about 
the  equal  angles  reciprocally  proportional :  and  trian- 
gles which  have  one  angle  in  the  one  equal  to  one  an- 
gle in  the  other,  and  their  sides  about  the  equal  angles 
reciprocally  proportional,  are  equal  to  one  another. 

Let  ABC,  ADE  be  equal  triangles,  which  have  the  angle  BAC 
equal  to  the  angle  DAE  ;  the  sides  about  the  equal  angles  of  the 
triangles  are  reciprocally  proportional ;  that  is,  CA  is  to  AD,  as 
EA  to  AB. 

Let  the  triangles  be  placed  so  that  their  sides  CA,  AD  be  in 
one  straight  line  ;  wherefore  also  EA  and  AB  are  in  one  straight 
line  a  ;  and  join  BD.     Because  the  triangle  ABC  is  equal  to  the  a  14. 1. 
triangle  ADE,  and  that  ABD  is 
another   triangle ;    therefore   as 
the  triangle  CAB  is  to  the  trian- 
gle BAD,  so  is  triangle  EAD  to 
triangle  DABb  :  but  as  triangle  /  ^'^S.  \         b  7- 5- 


CAB  to  triangle  BAD,  so  is  the 

base  C  A  to  AD  <=■ ;  and  as  trian-  /   y^  ^V^  \      c  1.  & 

gle  EAD  to  triangle  DAB,  so  is 

the  base  EA  to  AB^  ;  as  there-  C  E 

fore  C  A  to  AD,  so  is  E  A  to  AB  ^:  d  11. 5. 

wherefore  the  sides  of  the  triangles  ABC,  ADE  abaut  the  equal        ^ 

angles  are  reciprocally  proportional. 

But  let  the  sides  of  the  triangles  ABC,  ADE  about  the  equal 
angles  be  reciprocally  proportional,  viz.  C  A  to  AD,  as  EA  to  AB ; 
the  triangle  ABC  is  equal  to  the  triangle  ADB. 

Having  joined  BD  as  before  ;  because  as  CA  to  AD,  so  is  EA 
to  AB ;  and  as  CA  to  AD,  so  is  triangle  BAC  to  triangle ^AD^; 
and  as  EA  to  AB,  so  is  triangle  EAD  to  triangle  BAD  = ;  there- 
fore d  as  triangle  BAC  to  triangle  BAD,  so  is  triangle  EAD  to 
triangle  BAD ;  that  is,  the  triangles  BAC,  EAD  have  the  same 
ratio  to  the  triangle  BAD  :  wherefore  the  triangle  ABC  is  equal «  g  9.  5, 
to  the  triangle  ADE.     Therefore,  equal  triangles,  &c.     Q.  E.  D. 


172  THE  ELEMENTS 

Book  VI. 

PROP.  XVI.     THEOR. 


IF  four  straight  lines  be  proportionals,  the  rectangle 
contained  by  the  extremes  is  equal  to  the  rectangle 
contained  by  the  means  :  and  if  the  rectangle  contain- 
ed by  the  extremes  be  equal  to  the  rectangle  contained 
by  the  means,  the  four  straight  lines  are  proportionals. 


Let  the  four  straight  lines  AB,  CD,  E,  F  be  proportionals,  viz. 
as  AB  to  CD,  so  E  to  F ;  the  rectangle  contained  by  AB,  F  is 
equal  to  the  rectangle  contained  by  CD,  E. 

*  11. 1.  From  the  points  A,  C  draw  a  AG,  CH  at  right  angles  to  AB, 
CD  ;  and  make  AG  equal  to  F,  and  CH  equal  to  E,  and  com- 
plete the  parallelograms  BG,  DH :  because  as  AB  to  CD,  so  is 

h7.5.  E  to  F ;  and  that  E  is  equal  to  CH,  and  F  to  AG  ;  AB  is  •>  to 
CD,  as  CH  to  AG :  therefore  the  sides  of  the  parallelograms 
BG.  DH  about  the  e(|ual  angles  are  reciprocally  proportional  ; 
but   parallelograms  which  have  their  sides  about  equal  anglesr 

cl4.  6.  reciprocally  proportional,  are  equal  to  one  another*:  ;  therefore 
the  parallelogram  BG  is  equal  to  the  parallelogram  DH  :  and  the 
parallelogram  BG  is  contained      p 

i'  by  the  straight  lines  AB,  F,  be-  jj 

cause  AG  is  equal  to  F  ;  and 
the  parallelogram  DH  is  con- 
tained by  CD  and  E,  because 
CH  is  equal  to  E:  therefore 
the  rectangle  contained  by  the 
straight  lines  AB,  F  is  equal 
to  that  which  is  contained  by 
CD  and  E. 


And  if  the  rectangle  contain-  A  BCD 

ed  by*the  straight  lines  AB,  F 

be  equal  to  that  which  is  contained  by  CD,  E  ;  these  four  lines 
are  proportionals,  viz.  AB  is  to  CD,  as  E  to  F. 

The  same  construction  being  made,  because  the  rectangle 
contained  by  the  straight  lines  AB,  F  is  equal  to  that  which  is 
contained  by  CD,  E,  and  that  the  rectangle  BG  is  contained  by 
Ali,  F,  because  AG  is  equal  to  F;  and  the  rectangle  DH 
by  CD,  E,  because  CH  is  equal  to  E  ;  therefore  the  parallelogram 
BG  is  equal  to  the  parallelogram  DH ;  and  they  are  equiangu- 


OF  EUCLID. 


ifS 


lar:  but  the  sides  about  the  equal  angles  of  equal  parallelograms  Book  VI. 
are  reciprocally  pro,portionalc  :  wherefore,  as  A  B  to  CD,  so  isCH  ^■— ■/•"ii^ 
to  AG  ;  and  CH  is  equal  to  E,  and  AG  to  F :  as  therefore  AB  is  c  14.  6. 
to  CD,  so  E  to  F.     Wherefore,  if  four,  Sec    Q.  E.  D. 


PROP.  XVII.    THEOR. 


IF  three  straight  lines  be  proportionals,  the  rectan- 
gle contained  by  the  extremes  is  equal  to  the  square 
of  the  mean:  and  if  the  rectangle  contained  by  the 
extremes  be  equal  to  the  square  of  the  mean,  the  three 
straight  lines  are  proportionals. 


Let  the  three  straight  lines  A,  B,  C  be  proportionals,  viz.  as  A 
to  B,  so  B  to  C  ;  the  rectangle  contained  by  A,  C  is  equal  to  the 
square  of  B. 

Take  D  equal  to  B  ;  and  because  as  A  to  B,  so  B  to  C,  and  that 
B  is  equal  to  D  ;  A  is^  to  B,  as  D  to  C  :  but  if  four  straight  lines  a  7.  5. 
be  proportionals,  the  rect- 
angle  contained  by   the     A 
extremes  is  equal  to  that     B 
which  is  contained  by  the     D 

means  ^ :    therefore   the     C j  |      5 15. 6. 

rectangle  contained  by  A, 

C  is  equal  to  that  con-  |  "I  C  D 

tained  by  B,  D.    But  the 

rectangle  contained  by  B, 

D  is  the  square  of  B  ;  be-  A  B 

cause  B  is  equal  to  D  : 

therefore  the  rectangle  contained  by  A,  C  is  equal  to  the  square 

of  B. 

And  if  the  rectangle  contained  by  A,  C  be  equal  to  the  square 
of  B  ;  A  is  to  B,  as  B  to  C. 

The  same  construction  being  made,  because  the  rectangle 
contained  by  A,  C  is  equal  to  the  square  of  B,  and  the  square 
of  B  is  equal  to  the  rectangle  contained  by  B,  D,  because  B  is 
equal  to  D ;  therefore  the  rectangle  contained  by  A,  C  is  equal 
to  that  contained  by  B,  D :  but  if  the  rectangle  contained  by 
the  extremes  be  equal  to  that  contained  by  the  means,  the  four 
straight  lines  are  proportionals  b;  therefore  A  is  to  B,  as  D  to 


[  C 

. I 


174 


THE  ELEMENTS 


Book  VI.  C;  but  B  is  equal  to  D;  wherefore  as  A  to  B,  so  B  to  C. 
<— >— '  fore,  if  three  sti-aight  lines,  Sec.     Q.  E.  D. 


There. 


PROP.  XVIII.     PROB. 


See  N. 


UPON  a  given  straight  line  to  describe  a  rectili- 
neal figure  similar,  and  similarly  situated  to  a  given 
rectilineal  figure. 


a  23.  1. 


b  i2.  1. 


4.6. 


d  22.  5. 


Let  AB  be  the  sjiven  straight  line,  and  CDEF  the  given  recti- 
lineal figure  of  four  sides;  it  is  required  upon  the  given  straight 
line  AB  to  describe  a  rectilineal  figure  similar,  and  similarly  situ- 
ated to  CDEF. 

Join  DF,  and  at  the  points  A,  B,  in  the  straight  line  AB, 
make^  the  angle  BAG  equal  to  the  angle  at  C,  and  the  angle 
ABG  equal  to  the  angle  CDF ;  therefore  the  remaining  angle 
CFD  is  equal  to  the  remaining  angle  AGB^*:  wherefore  the 
triangle  FCD  is  e- 
quiangular  to  the 
triangle  GAB :  a- 
gain,  at  the  points 
G,  B,  in  the  straight 
line  GB,  make^  the 
angle  BGH  equal  to 
the  angle  DFE,  and 
the  angle  GBH  e- 
qual  to  FDE;  there- 
fore the  remaining  angle  FED  is  equal  to  the  remaining  angle 
GHB,  and  the  triangle  FDE  equiangular  to  the  triangle  GBH  : 
then,  because  the  angle  AGB  is  equal  to  the  angle  CFD,  and 
BGH  to  DFE,  the  whole  angle  AGH  is  equal  to  the  whole 
'CFE:  for  the  same  reason,  the  angle  ABH  is  equal  to  the  angle 
CDE  ;  also  the  angle  at  A  i-s  equal  to  the  angle  at  C,  and  the 
angle  GHB  to  FED :  therefore  the  rectilineal  figure  ABHG  is 
equiangular  to  CDEF  :  but  likewise  these  figures  have  their  sides 
about  the  equal  angles  proportionals:  because  the  triangles  GAB, 
FCD  being  equiangular,  BA  is^  to  AG,  as  DC  to  CF ;  and 
because  AG  is  to  GB,  as  CF  to  FD ;  and  as  GB  to  GH,  so, 
by  reason  of  the  equiangular  triangles  BGH,  DFE,  is  FD  to 
FE ;  therefore,  ex  xrjuali'^,  AG  is  to  Gil,  as  CF  to  FE :  in 
the  same  manner  it  may  be  proved  that  AB  is  to  BH,  as  CD  to 
DE :  and  GH  is  to  liB,  as  FE  to  ED  c.     Wherefore,  becausy 


OF  EUCLID.  ^^5 

the  rectilineal  figures  ABHG,  CDEF  are  equiangular,  and  have  ^^"okVI. 
their  sides  about  the  equal  angles  proportionals,  they  are  similar  '^"^ 
to  one  another  e.  ^  l.def.e. 

Next,  Let  it  be  required  to  describe  upon  a  given  straight  line 
AB,  a  rectilineal  figure  similar,  and  similarly  situated  to  the  rec-* 
tilineal  figure  CDKEF. 

Join  DE,  and  upon  the  given  straight  line  AB  describe  the 
rectilineal  figure  ABHG  similar,  and  similarly  situated  to  the 
quadrilateral  figure  CDEF,  by  the  former  case;  and  at  the 
points  B,  H,  in  the  straight  line  BH,  make  the  angle  HBL  equal 
to  the  angle  EDK,  and  the  angle  BHL  equal  to  the  angle 
DEK ;  therefore  the  remaining  angle  at  K  is  equal  to  the  re- 
maining angle  at  L  :  and  because  the  figures  ABHG,  CDEF 
are  similar,  the  angle  GHB  is  equal  to  the  angle  FED,  and 
BHL  is  equal  to  DEK  ;  wherefore  the  whole  angle  GHL  is 
equal  to  the  whole  angle  FEK  :  for  the  same  reason  the  angle 
ABL  is  equal  to  the  angle  CDK :  therefore  the  five  sided 
figures  AGHLB,  CFEKD  are  equiangular;  and  because  the 
figures  AGHB,  CFED  are  similar,  GH  is  to  HB,  as  FE  to 
ED  ;  and  as  HB  to  HL,  so  is  ED  to  EK^  ;  therefore,  e:r  cEcjuali^f  ^ 
GH  is  to  HL,  as  FE  to  EK :  for  the  same  reason,  AB  is  to  \iL,  ^  ^^'  ^' 
as  CD  to  DK:  and  BL  is  to  LH,  as^  DK  to  KE,  because  the 
triangles  BLH,  DKE  are  equiangular  ;  therefore,  because  the 
five  sided  figures  AGHLB,  CFEKD  are  equiangular,  and  have 
their  sides  about  the  equal  angles  proportionals,  they  are  similar 
to  one  another :  and  in  the  same  manner  a  rectilineal  figure  of 
six  or  more  sides  may  be  described  upon  a  given  straight  line 
similar  to  one  given,  and  so  on.     Which  was  to  be  done. 


PROP.  XIX.     THEOR. 


SIMILAR  triangles  are  to  one  another  in  the  du- 
plicate ratio  of  their  homologous  sides. 

Let  ABC,  DEF  be  similar  triangles,  having  the  angle  B  equal 
to  the  angle  E,  and  let  AB  be  to  BC,  as  DE  to  EF,  so  that  the 
side  BC  is  homologous  to  EF  «  ;    the  triangle  ABC  has  to  the  *12.def.5. 
triangle  DEF  the  duplicate  ratio  of  that  which  BC  has  to  EF. 

Take  BG  a  third  proportional  to  BC,  EF^  so  that  BC   is  to  b  11.  6 
EF,  as  EF   to   BG,   and  join  GA  ;    then,  because  as  AB  to  BC, 
3©  DE  to  EF  ;  alternately  S   AB  is  to  DE,  as  BC  to  EF  :  but  c  16-  5. 


176  THE,  ELEMENTS 

Book  VI.  as  BC   to  EF,  so  is  EF  to  BG  ;  therefore  d,  as  AB  to  DE,  so  is 
*^-^— ^  EF  to  EG  :  wherefore  the  sides  of  the  triangles  ABG,  DEF, 
d  11.  5.     which  are  about  the  equal  angles,  are  reciprocally  proportional  : 
but  triangles  which  have  the  sides  about  two  equal  angles  reci- 
procally  proportional, 
are  equal    to    one  an-  A 

e  15.  6.  other  <=  :  therefore  the 
triangle  ABG  is  equal 
to  the  triangle  DEF : 
and  because  as  BC  is 
to  EF,  so  EF  to  BG  ; 
and  that  if  three 
straight  lines  be  pro- 
portionals, the  first  is 
flO.def.5.  saidf  to  have  to  the  third  the  duplicate  ratio  of  that  which  it  has 
to  the  second  ;  BC  therefore  has  to  BG  the  duplicate  ratio  of 
1.6.  t^^^^  which  BC  has  to  EF  :  but  as  BC  to  BG,  so  is  e  the  trian- 
*  "  -  gle  ABC  to  the  triangle  ABG.  Therefore  the  triangle  ABC  has 
to  the  triangle  ABG  the  duplicate  ratio  of  that  which  BC  has 
to  EF  :  but  the  triangle  ABG  is  equal  to  the  triangle  DEF  ; 
wherefore  also  the  triangle  ABC  has  to  the  triangle  DEF  the  du- 
plicate ratio  of  that  which  BC  has  to  FF.  Therefore,  similar  tri- 
angles, &c.     Q.  E.  D. 

Cor.  From  this  it  is  manifest,  that  if  three  straight  lines  be 
proportionals,  as  the  first  is  to  the  third,  so  is  any  triangle  upon 
the  first  to  a  similar,  and  similarly  described  triangle  upon  the 
second. 


PROP,  XX.    THEOR. 

SIMILAR  polygons  may  be  divided  into  the  same 
number  of  similar  triangles,  having  the  same  ratio  to 
one  another  that  the  polygons  have  ;  and  the  polygons 
have  to  one  another  the  duplicate  ratio  of  that  vi^hich 
their  homologous  sides  have. 

Let  ABCDE,  FGHKL  be  similar  polygons,  and  let  AB  be  the 
homologous  side  to  FG :  the  polygons  ABCDE,  FGHKL  may 
be  divided  into  the  same  number  of  similar  triangles,  whereof 
each  to  each  has  the  same  ratio  which  the  polygons  have;  and 
the  polygon  ABCDE  has  to  the  polygon  FGHKL  the  duplicate 
ratio  of  that  wliich  the  side  AB  has  to  the  side  FG. 

Join  BE.  EC,  GL,  LH  :  and  because  the  polygon  ABCDE  is 


OF  EUCLID.  177 

similar  to  the  polygon  FGHKL,  the  angle  BAE  is  equal  to  the  Book  VL 
angle  GFL  a,  and  BA  is  to  AE  as  GF  to  FL  a  :  wherefore,  be-  ^— ^— i^^ 
cause  the  triangles  ABE,  FGL  have  an  angle  in  one  equal  to  a  Idef.  6. 
an  angle  in  the  other,  and  their  sides  about  these  equal  angles 
proportionals,  the  triangle  ABE  is  equiangular  •»,  and  therefore  b  6.  6. 
similar  to  the  triangle  FGL  <= ;  wherefore  the  angle  ABE  is  c  4.  6, 
equal  to  the  angle  FGL  :  and,  because  the  polygons  are  simi- 
lar, the  whole  angle  ABC  is  equal  »  to  the  whole  angle  FGH ; 
therefore  the  remaining  angle  EBC  is  equal  to  the  remaining 
angle  LGH :  and  because  the  triangles  ABE,  FGL  are  similar, 
EB  is  to  BA,  as  LG  to  GF  * ;  and  also,  because  the  polygons 
are  similar,  AB  is  to  BC,  as  FG  to  GH  ^;  therefore,  ex  aqualid^d  22.  $. 
EB  is  to  BC,  as  LG  to  GH  ;  that  is,  the  sides  about  the  equal  an- 
gles EBC,  LGH  are  proportionals;  therefore''  the  triangle  EBC 
is  equiangular 
to  the  triangle 
LGH,   and  si- 
milar   to    it  c. 
For  the   same 
reason,  the  tri- 
angle       ECD 
likewise  is  si- 
milar to  the  tri- 
angle   LHK  : 

therefore  the  similar  polygons  ABCDE,  FGHKL  are  divided  into 
the  same  number  of  similar  triangles. 

Also  these  triangles  have,  each  to  each,  the  same  ratio  which 
the  polygons  have  to  one  another,  the  antecedents  being  ABE,        ^ 
EBC,  ECD,  and  the  consequents  FGL,  LGH,  LHK :  and  tiie  po- 
lygon ABCDE  has  to  the  polygon  FGHKL  the  duplicate  ratio  of 
that  which  the  side  AB  has  to  the  homologous  side  FG. 

Because  the  triangle  ABE  is  similar  to  the  triangle  FGL, 
ABE  has  to  FGL  the  duplicate  ratio*  of  that  which  the  side  BE  «  19. 6, 
has  to  the  side  GL :  for  the  same  reason,  the  triangle  BEC  has 
to  GLH  the  duplicate  ratio  of  that  which  BE  has  to  GL  :  there- 
fore, as  the  triangle  ABE  to  the  triangle  FGL,  so^  is  the  trian-f  11.  5. 
gle  BEC  to  the  triangle  GLH.  Again,  because  the  triangle  EBC 
is  similar  to  the  triangle  LGH,  EBC  has  to  LGH  the  duplicate 
ratio  of  that  which  the  side  EC  has  to  the  side  LH :  for  the  same 
reason,  the  triangle  ECD  has  to  the  triangle  LHK  the  duplicate 
ratio  of  that  which  EC  has  to  LH  :  as  therefore  the  triangle  EBC 
to  the  triangle  LGH,  so  is  *'  the  triangle  ECD  to  the  triangle 
LHK :  but  it  has  been  proved  that  the  triangle  EBC  is  likewise 
to  the  triangle  LGH,  as  the  triangle  ABE  to  the  triangle  FGL.  ^ 
Therefore,  as  the  triangle  ABE  is  to  the  triangle  FGL,  so  is  tri- 
angle EBC  to  triangle  LGH,  and  triangle  ECD  to  triangle  LHK  : 
and,  therefore,  as  one  of  the  antecedents  to  one  of  the  consequents, 

Z 


irg  THE  ELEMENTS 

Book  VI.  SO  are  all  the  antecedents  to  all  the  consequents  s.  Where-* 
*«— v-"-'  fore,  as  the  triangle  ABE  to  the  triangle  FGL,  so  is  the  polygon 
g  12.  5.  ABCDE  to  the  polygon  FGHKL :  but  the  triangle  ABE  has  to 
the  triangle  FGL  the  duplicate  ratio  of  that  which  the  side  AB 
has  to  the  homologous  side  FG.  Therefore  also  the  polygon 
ABCDE  has  to  the  polygon  FGHKL  the  duplicate  ratio  of  that 
■which  AB  has  to  the  homologous  side  FG.  Wherefore,  similar 
polygons,  Sec.     Q.  E.  D. 

CoR.  1.  In  like  manner,  it  may  be  proved,  that  similar  four 
sided  figures,  or  of  any  number  of  sides,  are  one  to  another  in  the 
duplicate  ratio  of  their  homologous  sides,  and  it  has  already  been 
proved  in  triangles.  Therefore,  universally,  similar  I'ectilineal 
figures  are  to  one  another  in  the  duplicate  ratio  of  their  homolo- 
gous sides. 

Cor.  2.  And  if  to  AB,  FG,  two  of  the  homologous  sides,  a 
h  10.  def.  third  proportional  M  be  taken,  AB  has  *»  to  M  the  duplicate  ratio 
5-         of  that  which  AB  has  to  FG  :  but  the  four  sided  figure  or  poly- 
gon upon  AB  has  to  the  four  sided  figure  or  polygon  upon  FG 
likewise  the  duplicate  ratio  of  that  which  AB  has  to  FG  ;  there- 
fore, as  AB  is  to  M,  so  is  the  figure  upon  ABtothe  figure  upon  FG, 
i Cor.  19.  which  was  also  proved  in  triangles'.     Therefore,  universally,  it 
6.         is  manifest,  that  if  three  straight  lines  be  proportionals,  as  the 
first  is  to  the  third,  so  is  any  rectilineal  figure  upon  the  first,  to  a 
similar  and  similarly  described  rectilineal  figure  upon  the  second* 


OF  EUCLID. 


PROP.  XXI.    THEOR. 

.  RECTILINEAL  figures  which  are  similar  to  the 
same  rectilineal  figure,  are  also  similar  to  one  ano- 
ther. 

Let  each  of  the  rectilineal  figures  A,  B  be  similar  to  the  recti- 
lineal figure  C  :  the  figure  A  is  similar  to  the  figure  B. 

Because  A  is  similar  to  C,  they  are  equiangular,  and  also  have 
their  sides  about  the  equal  angles  proportionals*.  Again,  be- al.def.6., 
cause  B  is  similar  to 
C,they  are  equiangu- 
lar, and  have  their 
sides  about  the  equal 
angles  proportion- 
als * :  therefore  the 
figures  A,  B  are  each 

of  them  equiangular  to  C,  and  have  the  sides  about  the  equal  an- 
gles of  each  qf  them  and  of  C  proportionals.     Wherefore  the 
rectilineal  figures  A  and  B  are  equiangular  *>,  and  have  their  sides  b  1   Ax. 
about  the  equal  angles  proportionals  <=.     Therefore  A  is  similar  *   ^• 
to  B.    Q.  E.  D.  c  11.  5. 


PROP.  XXII.     THEOR. 

IF  four  straight  lines  be  proportionals,  the  similar 
rectilineal  figures  similarly  described  upon  them  shall 
also  be  proportionals ;  and  if  the  similar  rectilineal 
figures  similarly  described  upon  four  straight  lines 
be  proportionals,  those  straight  lines  shall  be  propor- 
tionals. 

Let  the  four  straight  lines  AB,  CD,  EF,  GH  be  proportionals, 
viz.  AB  to  CD,  as  EF  to  GH,  and  upon  AB,  CD  let  the  similar 
rectilineal  figures  KAB,  LCD  be  similarly  described ;  and  upon 
EF,  GH  the  similar  rectilineal  figures  MF,  NH  in  like  manner  : 
the  rectilineal  figure  KAB  is  to  LCD,  as  MF  to  NH. 

To  AB,  CD  take  a  third  proportional »  X  ;  and  to  EF,  GH  a  11.  6. 
a  third  proportional  O:  and  because  AB  is  to  CD,  as  EF  to  b  11.  5. 
GH,  and  that  CD  is  ^  to  X,  as  GH  to  O  ;  wherefore,  ejc  ac/iialic,  c  22.  5. 
as  AB  to  X,  so  EF  to  O :  but  as  AB  to  X,  so  is  ^  the  rectilineal  d  2.  Cor. 

20.  6. 


1«©  THE  ELEMENTS 

Book  VI.  KAB  to  the  rectilineal  LCD,  and  as  EF  to  O,  so  is<l  the  rectili* 
neal  MF  to  the  rectilineal  NH  :  therefore,  as  KAB  to  LCD,  so  >» 
is  MF  to  NH. 

And  if  the  rectilineal  KAB  he  to  LCD,  as  MF  to  NH  j  the 
straight  line  AB  is  to  CD,  as  EF  to  GH. 

Make «  as  AB  to  CD,  so  EF  to  PR,  and  upon  PR  describe  f 
the  rectilineal  figure  SR  similar  and  similarly  situated  to  either 


> 


M 

S 


E  F      G         H  PR 

of  the  figures  MF,  NH  :  then  because  as  AB  to  CD,  so  is  EF 
to  PR,  and  that  upon  AB,  CD  are  described  the  similar  and  si- 
milarly situated  rectilineals  KAB,  LCD,  and  upon  EF,  PR,  in 
like  manner,  the  similar  rectilineals  MF,  SR ;  KAB  is  to  LCD, 
as  MF  to  SR ;  but,  by  the  hypothesis,  KAB  is  to  LCD,  as  MF 
to  NH  ;  and  therefore  the  rectilineal  MF  having  the  same  ratio 
S^.S.  to  each  of  the  two  NH,  SR,  these  are  equals  to  one  another: 
they  are  also  similar,  and  similarly  situated  ;  therefore  GH  is 
equal  to  PR:  and  because  as  AB  to  CD,  so  is  EF  to  PR,  and 
that  PR  is  equal  to  GH  ;  AB  is  to  CD,  as  EF  to  GH.  If,  there- 
fore, four  straight  lines,  Sec.     Q.  E.  D. 


PROP.  XXHL     THEOR. 


Se,eN.  EQUIANGULAR  parallelograms  have  to  one 
another  the  ratio  which  is  compounded  of  the  ratios 
of  their  sides. 

Let  AC  CF  be  equiangular  parallelograms,  having  the  angle 
BCD  equal  to  the  angle  ECG :  the  ratio  of  the  parallelogram 
AC  to  the  parallelogram  CF  is  the  same  with  the  ratio  which  is 
com])ounded  of  the  ratios  of  their  sides. 


OF  EUCLID.  181 

Let  BC,  CG  be  placed  in  a  straight  line  ;  therefore  DC  and  Book  VI. 
CE  are  also  in  a  straight  line*  ;  and  complete  the  parallelogram  ^— v— ^ 
DG  ;  and,  taking  any  straight  line  K,  make^  as  BC  to  CG,  a  14. 1. 
so  K  to  L,  and  as  DC  to  CE,  so  make  ^  L  to  M  :  therefore  b  12.  6. 
the  ratios  of  K  to  L,  and  L  to  M,  are  the  same  with  the  ratios 
of  the  sides,  viz.  of  BC  to  CG,  and  DC  to  CE.     But  the  ra- 
tio of  K  to  M  is  that  which  is  said  to  be  compounded  c  of  thecA.dcf.5. 
ratios  of  K  to  L,  and  L  to  M  :   wherefore  also  K  has  to  M  the 

ratio  compounded  of  the  ratios  of  the     A D       H 

sides ;  and  because  as  BC  to  CG,  so  is 
the  parallelogram  AC  to  the  parallelo- 
gram CH  ^  ;  but  as  BC  to  CG,  so  is  K 
to  L  ;  therefore  K  is  ^  to  L,  as  the  pa- 
rallelogram AC  to  the  parallelogram 
CH  :  again,  because  as  DC  to  CE,  so 
is  the  parallelogram  CH  to  the  paral- 
lelogram CF  ;  but  as  DC  to  CE,  so  is 
L  to  M  ;  wherefore  L  is  e  to  M,  as  the 
parallelogram  CH  to  the  parallelogram 
CF  :  therefore,  since  it  has  been  proved 
that  as  K  to  L,  so  is  the  parallelogram  AC  to  the  parallelogram 
CH  ;  and  as  L  to  M,  so  the  parallelogram  CH  to  the  parallelo- 
,gram  CF  ;  ex  aquali  f,  K  is  to  M,  as  the  parallelogram  AC  to  the  f  22.  4, 
parallelogram  CF :  but  K  has  to  M  the  ratio  which  is  compound- 
ed of  the  ratios  of  the  sides  ;  therefore  also  the  parallelogram 
AC  has  to  the  parallelogram  CF  the  ratio  which  is  compounded 
of  the  ratios  of  the  sides.  Wherefore,  equiangular  parallelo- 
grams, &c.    Q.  E.  D. 


PROP.  XXIV.    THEOR* 


THE  parallelograms   about   the  diameter  of  any  see  n. 
parallelogram,  are  similar  to  the  whole,  and  to  one 
another. 


Let  ABCD  be  a  parallelog^ram,  of  which  the  diameter  is  AC  ; 
and  EG,  HK  the  parallelograms  about  the  diameter:  the  paral- 
lelograms EG,  HK  are  similar  both  to  the  whole  parallelogram 
ABCD,  and  to  one  another. 

Because  DC,  GF  are  parallels,  the  angle  ADC  is  equal  »  to  a  29. 1. 
the  angle  AGF :  for  the  same  reason,  because  BC,  EF  are  pa- 


182  THE  ELEMENTS 

BookVI.  rallels,  the  angle  ABC  is  equal  to  the  angle  AEF  :  and  each 

^— -V— -^  ol"  the  angles  BCD,  EFG  is  equal  to  the  opposite  angle  DAB  •>, 

b  34.  1.  and  therefore  are  equal  to  one  another ;  wherefore  the  paral- 
lelograms ABCD,  AEFG  are  equiangular :  and  because  the 
angle  ABC  is  equal  to  the  angle  AEF,  and  the  angle  BAC 
common  to  the  two  triangles  BAC,  EAF,  they  are  equiangu- 

C.4  6.  lar  to  one  another;  therefore <=  as  AB 
to  BC,  so  is  AE  to  EF  :  and  because 
the  opposite  sides  of  parallelograms 

d  7..  5-  are  equal  to  one  another '',  AB  is  ^  to 
AD,  as  AE  to  AG ;  and  DC  to  CB, 
as  GF  to  FE  ;  and  also  CD  to  DA, 
as  FG  to  GA :  therefore  the  sides  of 
the  parallelograms  ABCD,  AEFG 
about  the  equal  angles  are  proportion- 
als ;  and  they  are  therefore  similar  t© 

el.def.6.  one  another  <= :  for  the  same  reason,  the  parallelogram  ABCD 
is  similar  to  the  parallelogram  FHCK.  Wherefore  each  of  the 
parallelograms  GE,  KH  is  similar  to  DB  :  but  rectilineal  figures 
which  are  similar  to  the  same  rectilineal  figure,  ar^  also  similar 

£21.6.  to  one  another  f  ;  therefore  the  parallelogram  GE  is  similar  to 
KH.     Wherefore,  the  parallelograms,  Stc     Q.  E.  D. 


PROP.  XXV.    PROB. 


See  N.  'YO  describe  a  rectilineal  figure  which  shall  be  si- 
milar to  one,  and  equal  to  another  given  rectilineal 
figure. 

Let  ABC  be  the  given  rectilineal  figure,  to  which  the  figure  to 
be  described  is  required  to  be  similar,  and  D  that  to  which  it  must 
be  ecjual.     It  is  required  to  describe  a  rectilineal   figure  similar 
to  AIJC,  and  equal  to  D. 
a  Cor.  45.      Upon  the  straight   line  BC   describe^    the  parallelogram  BE 
^-         equal   to   the   figure   ABC  ;  also  upon  CE  describe  *   the  paral- 
lelogram CM  equal  to  D,  and   having    the    angle   FCE    equal 
,,^29.  Lio  the    angle    CBL  :    therefore    BC    and  CF  are  in  a   straight 
l^"*-  1- line  I',  as  also  LE  and  EM  :  between  BC  and  CF  find  <=  a  mean 
c  13.  6.     proportional    GH,    and    upon   CiH   describe 'l  the   rectilineal  fi- 
d  18.  6.     gure  KGII   similar   and   similarly  situated  to  the  figUVe  ABC  : 
e  2.  Cor.  and   because  BC  is  GH   as   GH  to  CF,  and   if  three   straight 
20.  (i.     lines  be  proportionals,    as  the  first  is  to  the  third,  so  is  «  the 


OF  EUCLID. 


18; 


figure  upon  the  first  to  the  similar  and  similarly  described  figure  Book  Vt 
upon  the  second ;  therefore  as  BC  to  CF,  so  is  the   rectilineal  ^ — v— ^ 
figure  ABC  to  KGH  :  but  as  BC  to  CF,  so  is  f  the  parallelogram  f  i.  6. 
BE  to  the  parallelogram  EF :  therefore  as  the  rectilineal  figure 
ABC  is  to  KGH,  so  is  the  parallelogi'am  BE  to  the  parallelogram 
EF  g :  and  the  rectilineal  figure  ABC  is  equal  to  the  parallelo-  g  11.  5. 


gram  BE ;  therefore  the  rectilineal  figure  KGH  is  equal  ^  to  the  h  14  5. 
parallelogram  EF :  but  EF  is  equal  to  the  figure  D  ;    wherefore 
also  KGH  is  equal  to  D  ;  and  it  is  similar  to  ABC.     Therefore 
the  rectilineal  figure  KGH  has  been  described  similar  to  the 
figure  ABC,  and  equal  to  D.     Which  was  to  be  done.  • 


PROP.  XXVI.    THEOR. 

IF  two  similar  parallelograms  have  a  common  an- 
gle, and  be  similarly  situated;  they  are  about  the  same 
diameter. 


Let  the  parallelograms  ABCD,  AEFG  be  similar  and  simi- 
larly situated  and  have  the  angle  DAB  common.  ABCD  and 
AEFG  are  about  the  same  diameter. 

For,  if  not,  let,  if  possible,  the 
parallelogram  BD  have  its  diame- 
ter AHC  in  a  different  straight 
line  from  AF,  the  diameter  of  the 
parallelogram  EG,  and  let  GF 
meet  AHC  in  H ;  and  through  H 
draw  HK  parallel  to  AD  or  BC : 
therefore  the  parallelograms 
ABCD,  AKHG  being  about  the 
same  diameter,  they  are  similar  to  one  another, »:  wherefore,  as  a  24.  6. 
DA  to  AB,  so  isb  GA  to  AK:  but  because  ABCD  and  AEFG 
are  similar  parallelograms,  as  DA  is  to  AB,  so  is  GA  to  AE;bl,def.6, 


184  THE  ELEMENTS 

Book  VI.  therefore  ^  as  GA  to  AE,  so  GA  to  AK ;  wherefore  GA  has  the 
same  ratio  to  each  of  the  straight  lines  AE,  AK ;  and  conse- 
quently AK  is  equal  ^  to  AE,  the  less  to  the  greater,  which 
is  impossible  :  therefore  ABCD  and  AKHG  are  not  about  the 
same  diameter;  wherefore  ABCD  and  AEFG  must  be  about 
the  same  diameter.     Therefore,  if  two  similar,  See.     Q.  E.  D. 

*  To  understand  the  three  following  propositions  more  easily, 
'  it  is  to  be  observed, 

'  1.  That  a  parallelogram  is  said  to  be  applied  to  a  straight 
'  line,  when  it  is  described  upon  it  as  one  of  its  sides.  JSx.  gr. 
'  the  parallelogram  AC  is  said  to  be  applied  to  the  straight  line* 
»  AB. 

'  2.  But  a  parallelogram  AE  is  said  to  be  applied  to  a  straight 
'  line  AB,  deficient  by  a  parallelogram,  when  AD  the  base  of 

*  AE  is  less  than  AB,  and  therefore 
'  AE  is  less  than  the  parallelogram 

*  AC  described  upon  AB  in  the  same 
'  angle,  and  between  the  same  pa- 
'  rallels,  by  the  parallelogram  DC  ; 

*  and  DC  is  therefore  called  the  de- 
'  feet  of  AE. 

*  3.  And  a  parallelogram  AG  is 
'  said  to  be  applied  to  a  straight  line  AB,  exceeding  by  a  parallel- 
'  ogram,  when  AF  the  base  of  AG  is  greater  than  AB,  and  there- 
'  fore  AG  exceeds  AC  the  parallelogram  described  upon  AB  in 
'  the  same  angle,  and  between  the  same  parallels,  by  the  paral- 
'  lelogram  BG.' 


PROP.  XXVII.    THEOR. 


SeeN. 


OF  all  parallelograms  applied  to  the  same  straight 
line,  and  deficient  by  parallelograms,  similar  and  si- 
milarly situated  to  that  which  is  described  upon  the 
lialf  of  the  line;  that  which  is  applied  to  the  half,  and 
is  similar  to  its  defect,  is  the  greatest. 

Let  AB  be  a  straight  line  divided  into  two  equal  parts  in  C ; 
and  let  the  parallelogram  AD  be  applied  to  the  half  AC ; 
which  is  therefore  deficient  from  the  parallelogram  upon  the 
whole  line  AB  by  the  parallelogram  CE  upon  the  other  half 
CB :    of  all  Uie  parallelograms  applied  to  any  other  parts  of 


OF  EUCLID.  185 

AB,  and  deficient  by  parallelograms  that  are  similar,  and  simi-BookVI. 
larly  situated  to  CE,  AD  is  the  greatest.  ^— 'v— -^ 

Let  AF  be  any  parallelogram  applied  to  AK,  any  other  part  of 
AB  than  the  half,  so  as  to  be  deficient  from  the  parallelogram  upon 
the  whole  line  AB  by  the  parallelogram  KH  similar,  and  similar- 
ly situated  to  CE  ;  AD  is  greater  than  AF. 

First,  let  AK  the  base  of  AF,  be  greater  than  AC  the  half  of 
AB  ;  and  because  CE  is  similar  to  the  D     L     E 

parallelogram  KH,  they  are  about  the  I  '^K    1         J 

same  diameter 3 :  draw  their  diameter  I  j\|p      I       8,26.6. 

DB,  and  complete   the   scheme :   be-      p    / J    \i | 

•  cause  the  parallelogram  CF  is  equal  *>  i  "  j      K        I   H  b  43.  X. 

to  FE,  add  KH  to  ijoth,  therefore  the  /  /      I  \     1 

whole  CH  is  equal  to  the  whole  KE :  /  /      /    \  I 

but  CH  is  equal c  to  CG,  because  the  | [_1__\  ^  "^^^  *• 

base    AC  is  equal    to   the  base  CB  j  a  C    K         TK 

therefore  CG  is  equal  to  KE  :  to  each 

of  these  add  CF  ;  then  the  whole  AF  is  equal  to  the  gnomon 
CHL :  therefore  CE,  or  the  parallelogram  AD,  is  greater  than 
the  parallelogram  AF. 

Next,  let  AK  the  base  of  AF,  be  less 
than  AC,  and,  the  same  construction  be- 
ing made,  the  parallelogram  DH  is  equal 
to  DG  c,  for  HM  is  equal  to  MG  d,  be- 
cause BC  is  equal  to  CA;  wherefore 
DH  is  greater  than  LG  :  but  DH  is 
equal  ^  to  DK  ;  therefore  DK  is  greater 
than  LG  :  to  each  of  these  add  AL  ;  then 
the  whole  AD  is  greater  than  the  whole 
AF.  Therefore,  of  all  parallelograms 
applied,  £cc.     Q.  E.  D. 

A       K     C 


2  A 


THE  ELEMENTS 


PROP.  XXVIII.    PROS. 


SeeN. 


a  10.  1. 


b  18.  6. 


c  25.  6. 
d  21.  6. 


TO  a  given  straight  line  to  apply  a  parallelogram 
equal  to  a  given  rectilineal  figure,  and  deficient  by  a 
parallelogram  similar  to  a  given  parallelogram  :  but 
the  given  rectilineal  figure  to  which  the  parallelogram 
to  be  applied  is  to  be  equal,  must  not  be  greater  than 
the  parallelogram  applied  to  half  of  the  given  line, 
having  its  defect  similar  to  the  defect  of  that  which 
is  to  be  applied  ;  that  is,  to  the  given  parallelogram. 

Let  AB  be  the  given  straight  line,  and  C  the  given  rectilineal 
figure,  to  which  the  parallelogram  to  be  applied  is  required  to 
be  equal,  which  figure  must  not  be  greater  than  the  parallelo- 
gram applied  to  the  half  of  the  line  having  its  defect  from  that 
upon  the  whole  line  similar  to  the  defect  of  that  which  is  to  be 
applied  ;  and  let  D  be  the  parallelogram  to  which  this  defect  is 
required  to  be  similar.  It  is  required  to  apply  a  parallelogram 
to  the  straight  line  AB,  which 
shall  be  equal  to  the  figure  C, 
and  be  deficient  from  the  pa- 
rallelogram upon  the  whole 
line  by  a  parallelogram  simi- 
lar to  D. 

^  Divide  AB  into  two  equal 
parts  *  in  t'le  point  E,  and 
upon  EB  describe  the  paral- 
lelogram EBFG  similar  b  and 
similarly  situated  to  D,  and 
complete  the  parallelogram 
AG,  which  must  either  be 
equal  to  C,  or  greater,  than  it. 
by  the  determination :  and  if 
AG  be  equal  to  C,  then  what  was  required  is  already  done  ; 
for,  upon  the  straight  line  AB,  the  parallelogram  AG  is  applied 
equal  to  tl:  ■  figure  C,  and  deficient  by  the  parallelogram  EF 
similar  to  D  :  but,  if  AG  be  net  equal  to  C,  it  is  greater  than 
it ;  and  EF  is  equal  to  AG  ;  therefore  EF  also  is  greater  than 
C.  Make  <=  the  parallelogr^im  KLMN  equal  to  the  excess  of 
EF  above  C,  and  similar  and  similarly  situated  to  D  ;  but  D  is 
similar  to  EF,  therefore  <i  also  KM  is  similar  to  EF  :  let  KL 


OF  EUCLID.  187 

be  the  homologous  side  to  EG,  and  LM  to  GF  :  and  because  Book  VI. 
EF  is  equal  to  C  and  KM  together,  EF  is  greater  than  KM  ;  ^  -,-._/ 
therefore  the  straight  line  EG  is  greater  than  KL,  and  GF  than 
LM  :  make  GX  equal  to  LK,  and  GO  equal  to  LM,  and  com- 
plete the  parallelogram  XGOP  :  therefore  XO  is  equal  and  si- 
milar to  KM ;  but  KM  is  similar  to  EF  ;  wherefore  also  XO  is 
similar  to  EF,  and  therefore  XO  and  EF  are  about  the  same  dia- 
meter e  :    let  GPB  be  their  diameter,  and  complete  the  scheme :  e  26.  6. 
then  because  EF  is  equal  to  C  and  KM  together,  and  XO  a  part 
of  the  one  is  equal  to  KM  a  part  of  the  other,  the  remainder, 
viz.  the  gnomon  ERO,  is  equal  to  the  remainder  C  :  and  because 
OR  is  equal  f  to  XS,  by  adding  SR  to  each,  the  whole  OB  is  £34. 1. 
equal  to  the  whole  XB  :  but  XB  is  equal  s  to  TE,  because  the  g  36. 1. 
base  AE  is  equal  to  the  base  EB ;  wherefore  also  TE  is  equal  to 
OB  :  add  XS  to  each,  then  the  whole  TS  is  equal  to  the  whole, 
viz.  to  the  gnomon  ERO  :  but  it  has  been  proved,  that  the  gno- 
mon ERO  is  equal  to  C,  and  therefore  also  TS  is  equal  to  C. 
Wherefore  the  parallelogram  TS,  equal  to  the  given  rectilineal 
iigure  C,  is  applied  to  the  given  straight  line  AB  deficient  by  the 
parallelogram  SR,  similar  to  the  given  one  D,  because  SR  is  si- 
milar to  EF  h.     Which  was  to  be  done.  h  24.  6. 


PROP.  XXIX.    PROB. 


TO  a  given  straight  line  to  apply  a  parallelogram  See  n. 
equal  to  a  given  rectilineal  figure,    exceeding  by  a 
parallelogram  similar  to  another  given. 


Let  AB  be  the  given  straight  line,  and  C  the  given  rectilineal 
figure  to  which  the  parallelogram  to  be  applied  is  required  to  be 
equal,  and  D  the  parallelogram  to  which  the  excess  of  the  one 
to  be  applied  above  that  upon  the  given  line  is  required  to  be  si- 
milar. It  is  required  to  apply  a  parallelogram  to  the  given 
straight  line  AB,  which  shall  be  equal  to  the  figure  C,  exceeding 
by  a  parallelogram  similar  to  D. 

Divide  AB  into  two  equal  parts  in  the  point  E,  and  upon  EB 
describe  »  the  parallelogram  EL  similar  and,  similarly  situa-  a  18.  6. 


188  THE  ELEMENTS 

Book  VI.  ted  to  D:  and  make  ^  the  parallelogram  GH  equal  to  EL  and 
s— yi^.1^  C  toj^ether,  and  similar  and  similarly  situated  to  D  ;  wherefore 
b  25.  6.  GH  is  similar  to  EL  <= :  let  KH  be  the  side  homologous  to  FL, 
c  21.  6,  and  KG  to  FE  :  and  because  the  parallelogram  GH  is  greater 
than  EL,  therefore  the  side  KH  is  greater  than  FL,  and  KG 
than  VE  :  produce  FL  and  FE,  and  make  FLM  equal  to  KH, 
and  FEN  to  KG,  and  complete  the  parallelogram  MN.  MN  is 
therefore  equal  and 
similar  to  GH  ;  but 
GH  is  similar  to  EL  ; 
wherefore  MN  is  si- 
milar to  EL,  and  con- 
sequently EL  and 
MN  are  about  the 
d  26.  6.  same  diameter  ^  : 
draw  their  diameter 
FX,  and  complete 
the  scheme.  There- 
fore, since  GH  is 
equal  to  EL  and  C 
together,  and  that 
GH  is  equal  to  MN ;  N  P         X 

MN  is  equal  to  EL  and  C  :  take  away  the  common  part  EL  ; 
then  the  remainder,  viz.  the  gnomon  NOL,  is  equal  to  C    And 
e  3S.  1.     because  AE  is  equal  to  EB,  the  parallelogram  AN  is  equal  «  to 
f  43. 1.     the  parallelogram  NB,  that  is,  to  BM  f.    Add  NO  to  each  :  there- 
fore the  whole,  viz.  the  parallelogram  AX,  is  equal  to  the  gno- 
mon NOL.     But  the  gnomon  NOL  is  equal  to  C  ;  therefore  also 
AX  is  equal  to  C.     Wherefore  to  the  straight  line  AB  there  is 
applied  the  parallelogram  AX  equal  to  the  given  rectilineal  C, 
exceeding  by  the  parallelogram  PO,  which  is  similar  to  D,  be- 
g  24. 6.    cause  PO  is  similar  to  EL  s.     Which  was  to  be  done. 


PROP.  XXX.     PROB. 


TO  cut  a  given  straight  line  in  extreme  and  mean 
ratio. 


Let  AB  be  the  given  straight  line  ;  it  is  required  to  cut  it  in 
extreme  and  mean  ratio. 


OF  EUCLID. 

Upon  AB  describe  »  the  square  BC,  and  to  AC  apply  the 
parallelogram  CD  equal  to  BC,  exceeding  by  the  figure  AD  si- 
milar to  BC  ^  :  but  BC  is  a  square, 
therefore  also  AD  is  a  square ;  and  be- 
cause BC  is  equal  to  CD,  by  taking  the 
common  part  CE  from  each,  the  re- 
mainder BF  is  equal  to  the  remainder  ^  t  IB 
AD  :  and  these  figures  ai'e  equiangular, 
therefore  their  sides  about  the  equal 
angles  are  reciprocally  proportional  <^ : 
wherefore,  as  FE  to  ED,  so  AE  to  EB  : 
but  FE  is  equal  to  AC^,  that  is,  to  AB; 
and  ED  is  equal  to  AE :  therefore,  as 
BA  to  AE,  so  is  AE  to  EB :  but  AB  is  "" 
greater  than  AE;  wherefore  AE  is  greater  than  EB^;  therefore 
the  straight  line  AB  is  cut  in  extreme  and  mean  ratio  in  E^. 
Which  was  to  be  done. 


189 


Book  VI. 


a  46. 1. 
b  29.  6. 


c  14.  6. 
d  34.  1. 


e  14.  5. 

f3.  dcf. 

6. 


Otherivise, 

Let  AB  be  the  given  straight  line ;   it  is  required  to  cut  it  in 
extreme  and  mean  ratio. 

Divide  AB  in  the  point  C,  so  that  the  rectangle  contained  by 

AB,  BC  be  equal  to  the  square  of  ACe.     Then, . g  11.  2. 

because  the  rectangle  AB,  BC  is  equal  to  the     .  '       „ 

square  of  AC,  as  BA  to  AC,  so  is  AC  to  CB^^ :    '  h  ir.  6. 


therefore  AB  is  cut  in  extreme  and  mean  ratio  in  C*". 
was  to  be  done. 


C 

Which 


PROP.  XXXL    THEOR. 


IN  right  angled  triangles,  the  rectilineal  figure  des-  See  n: 
cribed  upon  the  side  opposite  to  the  right  angle  is 
equal  to  the  similar  and  similarly  described  figures 
upon  the  sides  containing  the  right  angle. 

Let  ABC  be  a  right  angled  triangle,  having  the  right  angle 
BAC :  the  rectilineal  figure  described  upon  BC  is  equal  to  the 
similar  and  similarly  described  figures  upon  BA,  AC. 

Draw  the  perpendicular  AD  ;  therefore,  because  in  the  right 
angled  triangle  ABC,  AD  is  drawn  from  the  right  angle  at  A 
perpendicular  to  the  base  BC,  the  triangles  ABD,  ADC  are  si- 
milar-to  the  whole  triangle   ABC,  and  to  one  another »,  and  a  8.  6. 


!9ft 


THE  ELEMENTS 


20.6. 


dB.  5. 


BookVI.  because  the  triangle  ABC  is  similar  to  ADB,  as  CB  to  BA,  s© 
*— v^-'  is  BA  to  BD'';  and  because  these  three  straight  lines  are  pro- 
b  4.  6.  portionals,  as  the  first  to  the  third,  so  is  the  figure  upon  the  first 
c  2.  Cor.  to  the  similar  and  similarly  describ(>(j  figure  upon  the  second^: 
''^  "  therefore,  as  CB  to  BD,  so  is  the 
figure  upon  CB  to  the  similar  and 
similarly  described  figure  upon 
BA :  and,  inversely  ^,  as  DB  to 
BC,  so  is  the  figure  upon  BA  to 
that  upon  BC ;  for  the  same  rea- 
son, as  DC  to  CB,  so  is  the  figure 
uponCA  to  that  upon  CB.  Where- 
fore, as  BD  and  DC  together  to 
BC,  so  are  the  figures  upon  BA, 

AC  to  that  upon  BC^:  but  BD  and  DC  together  are  equal  to 
BC.  Therefore  the  figure  described  on  BC  is  equal  ^  to  the  si- 
milar and  similarly  described  figures  on  BA,  AC.  Wherefore, 
in  right  angled  triangles,  Sec.     Q.  E.  D. 


e24.  5 
f  A.  5. 


PROP.  XXXII.     THEOR. 


See  N.  IF  two  triangles  which  have  two  sides  of  the  one 
proportional  to  two  sides  of  the  other  be  joined  at 
one  angle,  so  as  to  have  their  homologous  sides  paral- 
lel to  one  another,  the  remaining  sides  shall  be  in  a 
straight  line. 


a  29. 1. 


Let  ABC,  DCE  be  two  triangles  which  have  the  two  sides 
BA,  AC  proportional  to  the  two  CD,  DE,  viz.  BA  to  AC  as  CD 
to  DE ;  and  let  AB  be  parallel  to  DC,  and  AC  to  DE.  BC  aiid 
CE  arc  in  a  straight  line. 

Because  AB  is  pirallel  to  A^ 
DC,  and  the  straight  line  AC 
meets  them,  the  alternate  an- 
gles BAC,  ACD  are  equal  »  ; 
for  the  same  reason,  the  angle 
CDE  is  equal  to  the  angle 
ACD  ;  wherefore  also  BAC  is 
equal  to  CDE :    and  because 


OF  EUCLID.  191 

tho,  triangles  ABC,  DCE  have  one  angle  at  A  equal  to  one  at  D,  Book  VI. 
and  the   sides  about    these    angles    proportionals,    viz.    BA   to  '^— v— ^ 
AC  as  CD  to  DE,  the  triangle' ABC  is  equiangular  b  to  DCE  :  b  6.  6. 
therefore  the  angle  ABC  is  equal  to  the  angle  DCE:  and  the 
angle   BAC   was  proved   to   be   equal   to   ACD  :    therefore   the 
whole  angle  ACE  is  equal  to  the  two  angles  ABC,  BAC  ;  add 
the  common  angle  ACB,  tiien  the  angles  ACE,  ACB  are  equal 
to  the  angles  ABC,  BAC,  ACB  :  but  ABC,  BAC,  ACB  are  equal 
to  two  right  angles  c;  therefore  also  the  angles  ACE,  ACB  are  c  32.  l-. 
equal  to  two  right  angles  :  and  since  at  tiie  point  C,  in  the  straight 
line  AC,  the  two  straight  lines  BC,  CE,  which  are  on  the  oppo- 
site sides  of  it,  make  the  adjacent  angles  ACE,  ACB  equal  to  two 
right   angles;    therefore '^  BC   and  CK   are   in  a  sl;'aight  line,  d  14.  l*. 
Wherefore,  if  two  triangles,  &c.     Q.  E.  D. 


PROP.  XXXIII.    THEOR. 


IN  equal  circles,  angles,  whether  at  the  centres  or  See  Nr. 
circumferences,  have  the  same  ratio  which  the  cir- 
cumferences on  which  they  stand  have  to  one  another : 
so  also  have  the  sectors. 


Let  ABC,  DEF  be  equal  circles  ;  and  at  their  centres  the 
angles  BGC,  EHF,  and 'the  angles  BAC,  EDF  at  their  cir- 
cumferences ;  as  the  circumference  BC  to  the  circumference 
EF,  so  is  the  angle  BGC  to  the  angle  EHF,  and  the  angle 
BAC  to  the  angle  EDF  ;  and  also  the  sector  BGC  to  the  sector 
EHF. 

I'ake  any  number  of  circumferences  CK,  KL,  each  equal  to 
BC,  and  any  number  whatever  FM,  MN  each  equal  to  EF  : 
and  join  GK,  GL,  HM,  HN.  Because  the  circumferences 
BC,  CK,  KL  are  all  equal,  the  angles  BGC,  CGK,  KGL 
are  also  all  equal*:  therefore  what  multiple  soever  the  circum- a  27. 
ference  BL  is  of  the  circumference  BC,  the  same  multiple  is 
the  angle  BGL  of  the  angle  BGC  :  for  the  same  reason,  what- 
ever multiple  the  circumference  EN  is  of  the  circumference 
EF,  the  same  multiple  is  the  angle  EHN  of  the  angle  EHF; 


193 


THE  ELEMENTS 


Book  VI.  and  if  the  circumference  BL  be  equal  to  the  circumferejjce 
C^y-»>  EN,  the  angle  BGL  is  also  equal  =»  to  the  angle  EHN  ;  and 
a  27.  3.  if  the  circumference  BL  be  greater  than  EN,  likewise  the  angle 
BGL  is  greater  than  EHN  ;  and  if  less,  less  :  there  being  then 
four  magnitudes,  the  two  circumferences  BC,  EF,  and  the 
two  angles  BGC,  EHF  ;  of  the  circumference  BC,  and  of  the 
angle  BGC,  have  been  taken  any  equimultiples  whatever,  viz. 
the  circumference  BL,  and  the  angle  BGL  ;  and  of  the  circum- 
ference EF,   and  of  the  angle  EHF,  any  equimultiples  what-^ 


E     r 


ever,  viz.  the  circumference  EN,  and  the  angle  EHN  :  and 
it  has  been  proved,  that,  if  the  circumference  BL  be  greater 
than  EN,  the  angle  BGL  is  greater  than  EHN ;  and  if 
equal,   equal ;  and  if  less,  less :  as  therefore  the  circumference 

b  5.def.5.  BC  to  the  circumference  EF,  so  ^  is  the  angle  BGC  to  the 
angle  EHF  :  but  as  the  angle  BGC  is  to  the  angle  EHF,  so  is 

c  15. 5.     '  the    angle    BAC    to    the    angle    EDF,    for  each  is  double   of 

d  20.  3.  each<i:  therefore,  as  the  circumference  BC  is  to  EF,  so  is  the 
angle  BGC  to  the  angle  EHF,  and  the  angle  BAC  to  the  angle 
EDF. 

Also,  as  the  circumference  BC  to  EF,  so  is  the  sector  BGC 
to  the  sector  EHF.  Join  BC,  CK,  and  in  the  circumferences 
BC,  CK  take  any  points  X,  O,  and  join  BX,  XC,  CO,  OK  : 
then,  because  in  the  triangles  GBC,  GCK  the  two  sides  BG, 
GC    are    equal    to    the    two    CG,    GK,   and    that    they   contain 

e  4. 1.  equal  angles  ;  the  ba^e  BC  is  equal  «  to  the  base  CK,  and  the 
triangle  GBC  to  the  triangle  GCK  :  and  because  the  circum- 
ference BC  is  equal  to  the  circumference  CK,  the  remaining 
part  of  the  whole  circumference  of  the  circle  ABC  is  equal  to 
the  remaining  part  of  the  whole  circumference  of  the  same 
circle  :    wherefore  the  angle  BXC  is  equal  to  the  angle  COK^j 

f  U.  def.  3f,(i  ^]^^  segment  BXC  is  therefore  similar  to  the  segment  COK*" , 


OF  EUCLID. 


193 


and  they  are  upon  equal  straight  lines  BC,  CK  :  but  similar  seg-  Book  VI. 
ments  of  circles  upon  equal  straight  lines  are  equals  to  one  ano-  ^-"-y^*^ 
ther  :  therefoi'e  the  segment  BXC  is  equal  to  the  segment  COK  :  g  24.  3- 
and  the  triangle  BGC  is  equal  to  the  triangle  CGK  ;  therefore 
the  whole,  the  sector  BGC,  is  equal  to  the   whole,  the   sector 
CGK  :  for  the  same  reason,  the  sector  KGL  is  equal  to  each  of 
the  sectors  BGC,  CGK :  in  the  same  manner,  the  sectors   EHF, 
FHM,  MHN  may  be   proved  equal  to  one  another :  therefore, 
what  multiples  soever  the  circumference   BL  is  of  the  circum- 
ference BC,  the  same  multiple  is  the  sector  BGL  of  the  sector 
,BGC :  for  the  same  reason,   whatever  multiple  the  circumfer- 
ence EN  is  of  EF,  the  same  multiple  is  the  sector  EH  :  of  the 
sector  EHF  :  and  if  the  circumference  BL  be  equal  to  EN,  the 


sector  BGL  is  equal  to  the  sector  EHN  ;  and  if  the  circumfer- 
ence BL  be  greater  than  EN,  the  sector  BGL  is  greater  than 
the  sector  EHN  ;  and  if  less,  less:  since,  then,  there  are  four 
magnitudes,  the  two  circumferences  BC,  EF,  and  the  two  sec- 
tors BGC,  EHF,  and  of  the  circumference  BC,  and  sector  BGC, 
the  circumference  BL  and  sector  BGL  are  any  equal  multiples 
whatever;  and  of  the  circumference  EF,  and  sector  EHF,  the 
circumference  EN,  and  sector  EHN,  are  any  equimultiples  what- 
ever ;  and  that  it  has  been  proved,  if  the  circumference  BL  be 
greater  than  EN,  the  sector  BGL  is  greater  than  the  sector  EHN  ; 
and  if  equal,  equal  ;  and  if  less,  less.  Therefore^,  as  the  cir-b5.dcf.5. 
cumference  BC  is  to  the  circumference  EF,  so  is  the  sector  BGC 
to  the  sector  EHF.    Wherefore,  in  equal  circles,  &c.    Q.  E.  D. 


3B 


THE  ELEMENTS 


PROP.  B.     THEOR. 


See  N.  IF  an  angle  of  a  triangle  be  bisected  by  a  straight 
line,  which  likewise  cuts  the  base  ;  the  rectangle  con- 
tained by  the  sides  of  the  triangle  is  equal  to  the  rect- 
angle contained  by  the  segments  of  the  base,  together 
with  the  square  of  the  straight  line  bisecting  the  an- 
gle. 

Let  ABC  be  a  triangle,  and  let  the  angle  BAC  be  bisected  by 
the  straight  line  AD  ;  the  rectangle  BA,  AC  is  equal  to  the  rect- 
angle BD,  DC,  together  with  the  square  of  AD. 

a  5. 4.  Describe   the  circle  »  ACB    about  the  triangle,  and  produce 

AD  to  the  circumference  in  E,  and 
join  EC  :  then  because  the  angle 
BAD  is  equal  to  the  angle  CAE,  and 

b21.3.  the  angle  ABD  to  the  angle*'  AEC, 
for  they  are  in  the  same  segment ; 
the  triangles  ABD,  AEC  are  equi- 
angular  to   one    another :    therefore 

c4. 6.  as  BA.  to  AD,  so  is^  EA  to  AC, 
and   consequently  the  rectangle  BA, 

d  16.  6.     AC   is   equal  ^  to  the  rectangle  EA, 

c3.  2.  AD,  that  is^,  to  the  rectangle  ED, 
DA,  together  with  the  square  of  AD  : 

£35.3.  but  the  rectangle  ED,  DA  is  equal  to  the  rectangle  f  BD,  DC. 
Therefore  the  rectangle  BA,  AC  is  equal  to  the  rectangle  BD, 
DC,  together  with  the  square  of  AD.  Wherefore,  if  an  ancle, 
Sec.    Q.  E.  D. 

PROP.  C.    THEOR. 

See  N.  IF  from  any  angle  of  a  triangle  a  straight  line  be 
drawn  perpendicular  to  the  base ;  the  rectangle  con- 
tained by  the  sides  of  the  triangle  is  equal  to  the  rect- 
angle contained  by  the  perpendicular  and  the  diame- 
ter of  the  circle  described  about  the  triangle. 

Let  ABC  be  a  triangle,  and  AD  the  perpendicular  from  the 
angle  A  to  the  base  BC ;  the  rectangle  BA,  AC  is  equal  to  the 
rectangle  contained  by  AD  and  the  diameter  of  the  circle  de- 
scribed about  the  triangle. 

J 


OF  EUCLID. 

Describe  *  the  circle  ACB  about  the 
triangle,  and  draw  its  diameter  AE, 
and  join  EC :  because  the  right  angle 
BDA  is  equal  •>  to  the  angle  ECA  in  a 
semicircle,  and  the  angle  ABD  to  the  B 
angle  AEC  in  the  same  segment  <= ;  the 
triangles  ABD,  AEC  are  equiangular; 
therefore,  as^  BA  to  AD,  so  is  EA  to 
AC ;  and  consequently  the  rectangle 
BA,  AC  is  equal  e  to  the  rectangle  EA, 
AD.  If,  therefore,  from  an  angle, 
&c.    Q.  E.  D. 


e  16.  6. 


PROP.  D.    THEOR. 


THE  rectangle  contained  by  the  diagonals  of  aSecN. 
quadrilateral  inscribed  in  a  circle  is  equal  to  both  the 
rectangles  contained  by  its  opposite  sides. 

Let  ABCD  be  any  quadrilateral  inscribed  in  a  circle,  and  join 
AC,  BD  ;  the  rectangle  contained  by  AC,  BD  is  equal  to  the  two 
rectangles  contained  by  AB,  CD,  and  by  AD,  -BC*. 

Make  the  angle  ABE  equal  to  the  angle  DBC ;    add  to  each 
of  these  the  common  angle  EBD,  then  the  angle  ABD  is  equal 
to  the  angle  EBC :    and  the  angle   BDA  is  equal  »  to  the  an- a  21.  3. 
gle  BCE,  because  they  are  in  the  same  segment ;    therefore 
the  triangle    ABD    is    equiangular  h  ^'•'^  "^"•^^ 

to  the  triangle  BCE :  wherefore  ^  as  /  ^^  ^  ^*  ^' 

BC  is  to  CE  so  is  BD  to  DA  ;    and 
consequently  the  rectangle  BC,  AD 
is  equal  c  to  the  rectangle  BD,  CE  : 
again,  because  the  angle  ABE  is  e- 
qual  to  the  angle  DBC,  and  the  an- 
gle a  BAE  to  the  angle  BDC,  the  tri- 
angle ABE  is  equiangular  to  the  tri- 
angle BCD :  as  therefore  BA  to  AE, 
so  is  BD  to  DC  ;  wherefore  the  rect- 
angle BA,  DC  is  equal  to  the  rect- 
angle BD,  AE  :  but  the  rectangle  BC,  AD  has  been  shown  equal 
to  the  rectangle  BD,  CE  ;  therefore  the  whole  rectangle  AC,  BD^  d  1.  2. 
is  equal  to  the  rectangle  AB,  DC,  together  with  the  rectangle 
AD,  BC.     Therefor^,  the  rectangle,  &c.     Q.  E.  D. 


c  16.  6. 


•  This  is  a  lemma  of  CI.  Ptolomjcus,  in  page  9.  of  his  KryttM  o-u»t*|<5- 


THE 

ELEMENTS  OF  EUCLID. 


BOOK  XL 


DEFINITIONS. 

I. 
Book  XI.  A  SOLID  is  that  which  hath  length,  breadth,  and  thickness. 

'^"■^^^  II. 

That  which  bounds  a  soUd  is  a  superficies. 

III. 

A  sti'aight  line  is  perpendicular  or  at  right  angles  to  a  plane, 
when  it  makes  right  angles  with  every  straight  line  meeting  it 
in  that  plane. 

IV. 

A  plane  is  perpendicular  to  a  plane  when  the  straight  lines  drawn 
in  one  of  the  planes  perpendicularly  to  the  common  section  of 
the  two  planes  are  perpendicular  to  the  other  plane. 

V. 

The  inclination  of  a  straight  line  to  a  plane  is  the  acute  angle 
contained  by  that  straight  line  and  another  drawn  from  the 
point  in  which  the  first  line  meets  the  plane,  to  the  point  in 
wliich  a  perpendicular  to  the  plane  drawn  from  any  point  of 
the  first  line  above  the  plane,  meets  the  same  plane. 

VI. 

The  inclination  of  a  plane  to  a  plane  is  the  acute  angle  contained 
by  two  straight  Ijnes  drawn  from  any  the  same  point  of  their 
common  section  at  right  angles  to  it,  one  upon  one  plane,  and 
the  other  upon  the  other  plane. 


OF  EUCLID.  197 

VII.  Book  XI, 

Two  planes  are  said  to  have  the  same,  or  a  like  inclination  to  one  ^-—v*^ 
another,  which  two  other  planes  have,  when  the  said  angles  of 
inclination  are  equal  to  one  another. 

VIII. 
Parallel  planes  are  such  which  do  not  meet  one  another  though 
produced. 

IX. 
A  solid  angle  is  that  which  is  made  by  the  meeting  of  more  than  See  N. 
two  plane  angles,  which  are   not  in  the  same  plane,  in  one 
point. 

X. 
'  The  tenth  definition  is  omitted  for  reasons  given  in  the  notes.'  See  N. 

Similar  solid  figures  are  such  as  have  all  their  solid  angles  equal, 
each  to  each,  and  which  are  contained  by  the  same  number  of 
similar  planes. 

XII. 

A  pyramid  is  a  solid  figure  contained  by  planes  that  are  consti- 
tuted betwixt  one  plane  and  one  point  above  it  in  which  they 
meet. 

XIII. 

A  prism  is  a  solid  figure  contained  by  plane  figures  of  which  two 
that  are  opposite  are  equal,  similar,  and  parallel  to  one  ano- 
ther ;  and  the  others  parallelograms. 

XIV. 

A  sphere  is  a  solid  figure  described  by  the  revolution  of  a  semi- 
circle about  its  diameter,  which  remains  unmoved. 

XV. 

The  axis  of  a  sphere  is  the  fixed  straight  line  about  which  the 
semicircle  revolves. 

XVI. 

The  centre  of  a  sphere  is  the  same  with  that  of  the  semicircle. 

XVII. 

The  diameter  of  a  sphere  is  any  straight  line  which  passes 
through  the  centre,  and  is  terminated  both  ways  by  the  super- 
ficies of  the  sphere. 

XVIII. 

A  cone  is  a  solid  figure  described  by  the  revolution  of  a  right 
angled  triangle  about  one  of  the  sides  containing  the  right 
angle,  which  side  remains  fixed. 

If  the  fixed  side  be  equal  to  the  other  side  containing   the  right  *' 

angle,  the  cone  is  called  a  right  angled  cone  ;  if  it  be  less 
than  the  other  side,  an  obtuse  angled,  and  if  greater,  an  acute 
angled  cone. 


THE  ELEMENTS 

XIX. 

The  axis  of  a  cone  is  the  fixed  straight  line  about  which  the  tri- 
angle revolves. 

XX. 

The  base  of  a  cone  is  the  circle  described  by  that  side  containing 
the  right  angle,  which  revolves. 

XXI. 

A  cylinder  is  a  solid  figure  described  by  the  revolution  of  a  right 
angted  parallelogram  about  one  of  its  sides,  which  remains 
fixed. 

XXII. 
The  axis  of  a  cylinder  is  the  fixed  straight  line  about  which  the 
pai'allelogram  revolves. 

XXIII. 
The  bases  of  a  cylinder  are  the  circles  described  by  the  two  re- 
volving opposite  sides  of  the  parallelogram. 

XXIV. 

Similar  cones  and  cylinders  are  those  which  have  their  axes  and 
the  diameters  of  their  bases  p:  oportionals. 

XXV. 

A  cube  is  a  solid  figure  contained  by  six  equal  squares. 

XXVI. 

A  tetrahedron  is  a  solid  figure  contained  by  four  equal  and  equi- 
lateral triangles. 

XXVII. 
An  octahedron  is  a  solid  figure  contained  by  eight  equal  and  equi- 
lateral triangles. 

XXVIII. 
A  dodecahedron  is  a  solid  figure  contained  by  twelve  equal  penta- 
gons which  are  equilateral  and  equiangular. 

XXIX. 

An  icosahedron  is  a  solid  figure  contained  by  twenty  equal  and 

equilateral  triangles. 

DEF.  A. 
A  parallelepiped  is  a  solid  figure  contained  by   six  quadrilateral 

figures,  whereof  every  opposite  two  are  parallel. 


OF  EUCLID* 


PROP.  I.  THEOR. 


ONE  part  of  a  straight  line  cannot  be  in  a  plane  sec  n. 
and  another  part  above  it. 

If  it  be  possible,  let  AB,  part  of  the  straight  line  ABC,  be  in 
the  plane,  and  the  part  BC  above  it  :  and  since  the  straight  line 
AB  is  in  the  plane,  it  can  be  pro- 
duced in  that  plane  :  let  it  be  pro- 
duced to  D:  and  let  any  plane  pass 
through  the  straight  line  AD,  and 
be  turned   about  it  until   it  pass 
through  the  point  C  ;  and  because  the  points  B,  C   are  in  this 
plane,  the  straight  line  BC  is  in  it »  :  therefore  there  are  two  a7.def.,l. 
straight  lines  ABC,  ABD  in  the  same  plane  that  have  a  common 
segment  AB,  which  is  impossible''.     Therefore,  one  part,  S^cbCor.ll. 
Q.  E.  D.  1. 


PROP.  II.     THEOR. 

TWO  Straight  lines  which  cut  one  another  are  in  see  N. 
one  plane,  and  three  straight  lines  which  meet  one 
another  are  in  one  plane. 

Let  two  straight  lines  AB,  CD  cut  one  another  in  E  ;  AB, 
CD  are  one  plane  :  and  three  straight  lines  EC,  CB,  BE  which 
meet  one  another,  are  in  one  plane. 

Let  any  plane  pass  through  the  straight 
line  EB,  and  let  the  plane  be  turned  about 
EB,  produced,  if  necessary,  until  it  pass 
through  the  point  C  :  then  because  the 
points  E,  C  are  in  this  plane,  the  straight 

line  EC  is  in  it^  :  for  the  same  reason,  the  /    \  a7.def,l. 

straight  line  BC  is  in  the  same  ;  and,  by 
the  hypothesis,  EB  is  in  it:  therefore  the 
three  straight  lines  EC,  CB,  BE  are  in  one 
plane  :  but  in  the  plane  in  which  EC,  EB 
are,  in  the  same  are  ^  CD,  AB  :  therefore, 

AB,  CD  are  in  one  plane.     Wherefore,  two  straight  lines.  Sec. 
Q.  E.  D. 


THE  ELEMENTS 


PROP.  III.     THEOR. 


See  N. 


IF  two  planes  cut  one  another,  their  common  sec 
tion  is  a  straight  line. 


Let  two  planes  AB,  BC,  cut  one  another,  and  let  the  line  DB 
be  their  common  section  :  DB  is  a  straight 
line  :  if  it  be  not,  from  the  point  D  to  B 
draw,  in  the  plane  AB,  the  straight  line 
DEB,  and  in  the  plane  BC  the  straight 
line  DFB:  then  two  straight  lines  DEB, 
DFB  have  the  same  extremities,  and  there- 
fore include  a  space  betwixt  them  ;  which 
a  10.  Ax.  is  impossible  * :  therefore  BD  the  common 
1-  section  of  the  planes  AB.   BC  cannot  but 

be  a  straight  line.      Wherefore,   if  two 
planes,  he.    Q.  E.  D. 


PROP.  IV.    THEOR. 


See  N. 


5ll5.  1. 

b  4.  3. 


c26.  1. 


IF  a  straight  line  stand  at  right  angles  to  each  of 
two  straight  lines  in  the  point  of  their  intersection,  it 
shall  also  be  at  right  angles  to  the  plane  which  passes 
through  them,  that  is,  to  the  plane  in  which  they  are. 

Let  the  straight  line  EF  stand  at  right  angles  to  each  of  the 
straight  lines  AB,  CD  in  E,  the  point  of  their  intersection:  EF 
is  also  at  right  angles  to  the  plane  passing  through  AB,  CD. 

Take  the  straight  lines  AE,  EB,  CE,  ED  all  equal  to  one  an- 
other;  and  through  E  draw,  in  the  plane  in  which  are  AB,  CD, 
any  straight  line  GEFI ;  and  join  AD,  CB ;  then,  from  any  point 
F  in  EF',  draw  FA,  FG,  FD,  EC,  FH,  FB :  and  because  the 
two  straight  lines  AE,  ED  are  equal  to  the  two  BE,  EC,  and 
that  they  contain  equal  angles  ^  AED,  BEC,  the  base  AD  is 
equal  ^  to  the  base  BC,  and  the  angle  DAE  to  the  angle  EBC  : 
and  the  angle  AEG  is  equal  to  the  angle  BEH^;  therefore 
the  triangles  AEG,  BEH  have  two  angles  of  one  equal  to 
two  angles  of  the  other,  each  to  each,  and  the  sides  AE,  EB, 
adjacent  to  the  equal  angles,  equal  to  one  another ;  where- 
fore they  shall  have  their  other  sides  equal  «= :  GE  is  therefore 


OF  EUCLID. 


301 


equal  to  EH,  and  AG  to  BH  :   and  because  AE  is  equal  to  EB,  Book  XI. 
and  FE  common  and  at  right  angles  to  them,   the  base  AF  is  v — y-— ; 
equal  ^  to  the  base  FB  ;  for  the  same  reason,  CF  is  equal  to  FD  :  b  4. 1. 
and  because  AD  is  equal  to  BC,  and  AF  to  FB,  the  two  sides 
FA,  AD  are  equal  to  the  two  FB,  BC, 
each  to  each  ;    and  the  base  DF  was 
proved  equal  to  the  base  FC ;  therefore 

the  angle  FAD  is  equal  <*  to  the  angle  ///  \\\  d  8. 1. 

FBC  :  again,  it  was  proved  that  GA  is 
equal  to  BH,  and  also  AF  to  FB  ;    FA, 
then,  and  AG  are  equal  to  FB  and  BH, 
and  the  angle   FAG   has  been   proved 
equal  to  the  angle  FBH  ;   therefore  the 
base  GF  is  equal ''  to  the  base  FH:  again, 
because  it  was  proved,  that  GE  is  equal 
to  EH,  and  EF  is  common  ;  GE,  EF  are 
equal  to  HE,  EF  ;  and  the  base  GF  is 
equal  to  the  base  FH  ;  therefore  the  an- 
gle GEF  is  equal  ^  to  the  angle  HEF  ;  and  consequently  each  of 
these  angles  is  a  right  ^  angle.      Therefore  tE  makes  right  an- e  10.  def . 
gles  with  GH,  that  is,  with  any  straight  line  drawn  through  E  in    1 
the  plane  passing  through  AB,  CD.     In  like  manner,  it  may  be 
proved,  that  FE  makes   right  angles   with   every   straight  line 
which  meets  it  in  that  plane.     But  a  straight  line  is  at  right  an- 
gles to  a  plane  when  it  makes  right  angles  with  every  straight 
line  which  meets  it  in  that  plane  f;  therefore  EF  is  at  right  an- f  3.  def. 
gles  to  the  plane  in  which  are  AB,  CD.    Wherefore,  if  a  straight   11 
line,  &c.     Q,  E.  D. 


PROP.  V.    THEOR. 


IF  three  straight  lines  meet  all  in  one  point,  and  a  See  n. 
straight  line  stands  at  right  angles  to  each  of  them  in 
that  point ;  these  three  straight  lines  are  in  one  and 
the  same  plane. 


.  Let  the  straight  line  AB  stand  at  right  angles  to  each  of  the 
straight  lines  BC,  BD,  BE,  in  B  the  point  where  they  meet ;  BC, 
BD,  BE  are  in  one  and  the  same  plane. 

If  not,  let,  if  it  be  possible,  BD  and  BE  be  in  one  plane,  and 
BC  be  above  it ;  and  let  a  plane  pass  through  AB,  BC,  the  com- 
mon flection  of  which  with  the  plane,  in  which  BD  and  BE  are, 


202 


THE  ELEMENTS 


Book  XL  shall  be  a  straight  »  line  ;  let  this  be  BF  :  therefore  the  three 
> — ^-mJ  straight  lines  AB,  BC,  BF  are  all  in  one  plane,  viz.  that  which 
a  3.  11.    passes  through  AB,  BC  ;  and  because  AB  stands  at  right  angles 

to  each  of  the  straight  lines  BD,  BE,  it  is  also  at  right  angles 

b  4. 11.    ''to  the  plane  passing  through  them  ;  and  therefore  makes  right 

c  f?.  def.    angles  =  with  every  straight  line  meet- 

11.  ing  it  in  that  plane  ;  but  BF  which  is 

in  that  plane  meets  it:    therefore  the 

angle  ABF  is  a  right  angle  ;    but  the 

angle  ABC,  by  the  hypothesis,  is  also 

a  right  angle  ;  therefore  the  angle  ABF 

is  equal  to  the  angle  ABC,    and  they 

are  both  in  the   same  plane,  which  is 

impossible  :  therefore  the  straight  line 

BC  is  not  above  the  plane  in  which  are 

BD    and    BE :     wherefore   the   three 

straight  lines  BC,  BD,  BE  are  in  one  and  the  same  plane. 

fore,  if  three  straight  lines,  &;c.     Q.  E.  D. 


There- 


PROP.  VI.     THEOR. 


IF  two  straight  lines  be  at  right  angles  to  the  same 
plane,  they  shall  be  parallel  to  one  another. 


n. 


Let  the  straight  lines  AB,  CD  be  at  right  angles  to  the  same 
plane  ;  AB  is  parallel  to  CD. 

Let  them  meet  the  plane  in  the  points  B,  D,  and  draw  the 
straight  line  BD,  to  which  draw  DE  at  right  angles,  in  the  same 
plane  ;  and  make  DE  equal  to  AB,  and  join 
BE,  AE,  AD.  Then,  because  AB  is  per- 
a  3.  def.  pendicular  to  the  plane,  it  shall  make  right » 
angles  with  every  straight  line  which  meets 
it,  and  is  in  that  plane  :  but  BD,  BE,  which 
are  in  that  plane,  do  each  of  them  meet  AB. 
Therefore  each  of  the  angles  ABD,  ABE  is 
a  rigiit  angle:  for  the  same  reason,  each  of 
the  angles  CDB,  CDE  is  a  right  angle  : 
and  because  AB  is  equal  to  DE,  and  BD 
common,  the  two  sides  AB,  BD  are  equal 
to  the  two  ED,  DB  ;  and  they  contain  right 

angles  ;  therefore  the  base  AD  is  equal  ^  to  the  base  BE  :  again, 
because  AB  is  equal  to  UE,  and  BE  to  AD  ;  AB,  BE  are  equal 


b4. 1. 


OF  EUCLia  203 

to  ED,  DA  ;  and,  in  the  triangles  ABE,  EDA,  tlfc  base  AE  is  Book XI, 
common  ;  therefore  the  angle  ABE  is  equal  «=  to  the  angle  EDA  :  *>— v— ♦ 
but  ABE  is  a  right  angle;  therefore  EDA  is  also  a  right  angle,  c  8. 1. 
and  ED  perpendicular  to  DA  :  but  it  is  also  perpendicular  to  eaqh 
of  the  two  BD,  DC :  wherefore  ED  is  at  right  angles  to  each  of 
the  three  straight  lines  BD,  DA,  DC  in  the  point  in  which  they 
meet :  therefore  these  three  straight  lines  are  all  in  the   same 
plane  **:  but  AB  is  in  the  plane  in  which  are  BD,  DA,  because  d  5. 11, 
any  three  straight  lines  which  meet  one  another  are  in  one  plane  e  :  e  2.  11. 
therefore  AB,  BD,  DC  are  in  one  plane :  and  each  of  the  angles 
ABD,  BDC  is  a  right  angle;  therefore  AB  is  parallel  f  to  CD.  £28,1. 
Wher«fore,  if  two  straight  lines.  Sec.     Q.  E.  D. 


PROP.  VII.    THEOR. 

IF  two  straight  lines  be  parallel,  the  straight  line  See  n. 
drawn  from  any  point  in  the  one  to  any  point  in  the 
other  is  in  the  same  plane  with  the  parallels. 

Let  AB,  CD  be  parallel  straight  lines,  and  take  any  point  E 
in  the  one,  and  the  point  F  in  the  other:  the  straight  line  which 
joins  E  and  F  is  in  the  same  plane  with  the  parallels. 

If  not,  let  it  be,  if  possible,  above  the  plane,  as  EGF  ;  and  in 
the  plane  ABCD  in  which  the  paral-  a  F  r 

lels  are,  draw  the  straight  line  EHF 
from  E  to  F;  and  since  EGF  also 
is  a  straight  line,  the  two  straight 


lines  EHF,  EGF  include  a  space  be-  ^ 

tween  them,  which  is  impossible  a,  \\  a  10. 

Therefore  the  straight  line  joining     - — ■ ^1 A*!,. 

the   points   E,  F  is  not  above  the  C  F  D 

plane  in  which  the  parallels  AB,  CD  are,  and  is  therefore  in  that 
plane.     Wherefore,  if  two  straight  lines,  8cc.     Q.  E.  D. 


PROP.  VIII.    THEOR. 

IF  two  straight  lines  be  parallel,  and  one  of  them  sce  n, 
is  at  right  angles  to  a  plane,  the  other  also  shall  be 
at  right  angles  to  the  same  plane. 


'504 


THE  ELEMENl  S 


Book  XI 


gr.ii. 


a  3. 
def.  11. 


b  29. 1. 


c41. 


d8. 1. 


e  4. 11. 

f3.  def. 
11 


Let  AB,  CD  be  two  parallel  straight  lines,  and  let  one  of  them 
AB  be  at  right  angles  to  a  plane  ;  the  other  CD  is  at  right  angles 
to  the  same  plane. 

Let  AB,  CD  meet  the  plane  in  the  points  B,  D,  and  join  BD : 
therefore  e  AB,  CD,  BD  are  in  one  plane.  In  the  plane  to  which 
AB  is  at  right  angles,  draw  DE  at  right  angles  to  BD,  and  make 
DE  equal  to  AB,  and  join  BE,  AE,  AD.  And  because  AB  is  per- 
pendicular to  the  plane,  it  is  perpendicular  to  every  straight  line 
which  meets  it,  and  is  in  that  plane  »  :  therefore  each  of  the  angles 
ABDj  ABE  is  a  right  angle :  and  because  the  straight  line  BD 
meets  the  parallel  straight  lines  AB,  CD,  the  angles  ABD,  CDB 
are  together  equal  i>  to  two  right  angles:  and  ABD  is  a  right  an- 
gle ;  therefore  also  CDB  is  a  right  angle,  and  CD  perpendicular 
to  BD  :  and  because  AB  is  equal  to  DE,  and  BD  common,  the 
two  AB,  BD,  are  equal  to  the  two  ED, 
DB,  and  the  angle  ABD  is  equal  to  the  a 
angle  EDB,  because  each  of  them  is  a  * 
right  angle ;  therefore  the  base  AD  is 
equal  =  to  the  base  BE  :  again,  because  AB 
is  equal  to  DE,  and  BE  to  AD  ;  the  two 
AB,  BE  are  equal  to  the  two  ED,  DA; 
and  the  base  AE  is  common  to  the  trian-  g 
gles  ABE,  EDA ;  wherefore  the  angle 
ABE  is  equal d  to  the  angle  EDA:  and 
ABE  is  a  right  angle  ;  and  therefore  EDA 
is  a  right  angle,  and  ED  perpendicular  to 
DA:  but  it  is  also  perpendicular  to  BD ; 
therefore    ED    is   perpendicular  e    to    the 

plane  which  passes  through  BD,  DA,  and  shall f  make  right  an- 
gles with  every  straight  line  meeting  it  in  that  plane :  but  DC  is 
in  the  plane  passing  through  BD,  DA,  because  all  three  are  in 
the  plane  in  which  are  the  parallels  AB,  CD:  wherefore  ED  is 
at  right  angles  to  DC ;  and  therefore  CD  is  at  right  angles  to 
DE:  but  CD  is  also  at  right  angles  to  DB  ;  CD  then  is  at  right 
angles  to  the  two  straight  lines  DE,  DB  in  the  point  of  their  in- 
tersection D  ;  and  therefore  is  at  right  angles  *  to  the  plane  pass- 
ing through  DE,  DB,  which  is  the  same  plane  to  which  AB  is  at 
right  angles.     Therefore,  if  two  straight  lines,  &c.     Q.  E.  D. 


OF  EUCLID. 


PROP.  IX.    THEOR. 


TWO  straight  lines  which  are  each  of  them  paral- 
lel to  the  same  straight  line,  and  not  in  the  same  plane 
with  it,  are  parallel  to  one  another. 

Let  AB,  CD  be  each  of  them  parallel  to  EF,  and  not  in  the 
same  plane  with  it ;  AB  shall  be  parallel  to  CD. 

In  EF  take  any  point  G,  from  which  draw,  in  the  plane  passing 
through  EF,  AB,  the  straight  line  GH  at  right  angles  to  EF ; 
and  in  the  plane  passing  through  EF,  CD,  draw  GK  at  right  an- 
gles to  the  same  EF.      And  be- 
cause EF  is  perpendicular  both  to 
GH  and  GK,  EF  is  perpendicu- 
lar »  to  the  plane  HGK  passing  \  a  4. 11. 
through  them  :  and  EF  is  parallel 
to  AB ;   therefore  AB  is  at  right 
angles  •>  to  the  plane  HGK.     For 
the  same  reason,  CD  is  likewise 
at  right  angles  to  the  plane  HGK. 
Therefore  AB,  CD  are  each  of 
them  at  right  angles  to  the  plane 

HGK.     But  if  two  straight  lines  be  at  right  angles  to  the  same 
plane,  they  shall  be  parallel  <=  to  one  another.     Therefore  AB  isc  6.  It. 
parallel  to  CD.     Wherefore,  two  straight  lines,  &c.     Q.  E.  D. 


PROP.  X.    THEOR. 

IF  two  straight  lines  meeting  one  another  be  paral- 
lel to  two  others  that  meet  one  another,  and  are  not 
in  the  same  plane  with  the  first  two,  the  first  two  and 
the  other  two  shall  contain  equal  angles. 

Let  the  two  straight  lines  AB,  BC  which  meet  one  another  be 
parallel  to  the  two  straight  lines  DE,  EF  that  meet  one  another, 
and  are  not  in  the  same  plane  with  AB,  BC.  The  angle  ABC  is 
equal  to  the  angle  DEF. 

Take  BA,  BC,  ED,  EF  all  equal  to  one  another ;  and  join  AD, 
CF,  BE,  AC,  DF :  because  B  A  is  equal  and  parallel  to  ED,  there- 


206 


THE  ELEMENTS 


Book  XI.  fore  AD  is  *  both  equal  and  parallel  to  BE. 

^■— v—^  For  the  same  reason  CF  is  equal  and  pa- 

a  33. 1,  rallel  to  BE.  Therefore  AD  and  CF  are 
each  of  them  equal  and  parallel  to  BE. 
But  straight  lines  that  are  parallel  to  the 
same  straight  line,  and  not  in  the  same 

b  9. 11.  plane  with  it,  are  parallel  ''  to  one  ano- 
ther.    Therefore  AD  is  parallel  to  CF ; 

el.Ax.l.  and  it  is  equal  ^  to  it,  and  AC,  DF  join 
them  towards  the  same  parts  ;  and  there- 
fore a  AC  is  equal  and  parallel  to  DF. 
And  because  AB,  BC  are  equal  to  DE, 
EF,  and  the  base  AC  to  the  base  DF ;  the 
angle  ABC  is  equal «!  to  the  angle  DEF. 
straight  lines,  &c.     Q.  E.  D. 


d8. 1. 


Therefore,  if  two 


PROP.  XI.    PROB. 


TO  draw  a  straight  line  perpendicular  to  a  plane, 
from  a  given  point  above  it. 


a  12.  1. 


b  11. 1. 


c31. 1. 


d  4.  11. 


e  8. 11. 


P3.  def. 
11. 


Let  A  be  the  given  point  above  the  plane  BH ;  it  is  required 
to  draw  from  the  point  A  a  straight  line  perpendicular  to  the 
plane  BH. 

In  the  plane  draw  any  straight  line  BC,  and  from  the  point  A 
draw  a  AD  perpendicular  to  BC.  If  then  AD  be  also  perpendi- 
cular to  the  plane  BH,  the  thing  required  is  already  done ;  but  if 
it  be  not,  from  the  point  D  draw*",  in  the  plane  BH,  the  straight 
line  DE  at  right  angles  to  BC  :  and  from  the  point  A  draw  AF 
perpendicular  to  DE  >  and  through  F  draw  <=  GH  parallel  to  BC : 
and  because  BC  is  at  right  angles 
to  ED  and  DA,  BC  is  at  right  an- 
gles ^  to  the  plane  passing  through 
ED,  DA.  And  GH  is  parallel  to 
BC  ;  but,  if  two  straight  lines  be 
parallel, tJne  of  which  is  at  right 
angles  to  a  plane,  the  other  shall 
be  at  right  e  angles  to  the  same 
plane;  wherefore  GH  is  at  right 
angles  to  the  plane  through  ED, 
DA,  and  is  perpendicular  ^  to 
every  straight  line  meeting  it  in  that  plane.  But  AF,  which  is 
in  the  plane  through  ED,  DA,  meets  it:    therefore  GH  is  pgK- 


OF  EUCLID. 


207 


pendicular  to  AF  ;  and  consequently  AF  is  perpendicular  to  GH ;  Book  XL 
and  AF  is  perpendicular  to  DE:  therefore  AF  is  perpendicular  *-.,-,^v_f 
to  each  of  the  straight  lines  GH,  DE.  But  if  a  straight  jine  stands 
at  right  angles  to  each  of  two  straight  lines  in  the  point  of  their 
intersection,  it  shall  also  be  at  right  angles  to  the  plane  passing 
through  them.  But  the  plane  passing  through  ED,  GH  is  the 
plane  BH  ;  therefore  AF  is  perpendicular  to  the  plane  BH  ;  there- 
fore, from  the  given  point  A,  above  the  plane  BH,  the  straight 
line  AF  is  drawn  perpendicular  to  that  plane.  Which  was  to  be 
doneu 


PROP.  XH.     PROB. 


TO  erect  a  straight  line  at  right  angles  to  a  given 
plane,  from  a  point  given  in  the  plane. 


D     B 


Let  A  be  the  point  given  in  the  plane  ;  it  is  required  to  erect 
a  straight  line  from  the  point  A  at  right 
angles  to  the  plane. 

From  any  point  B  above  the  plane  draw  » 
BC  perpendicular  to  it ;  andfrora  A  draw*" 
AD  parallel  to  BC.  Because,  tiierefore, 
AD,  CB  are  two  parallel  straight  lines,  r 

and  one  of  them  BC  is  at  right  angles  to        / 
the  given  plane,  the  other  AD  is  also  at       / 

right  angles  to  it  <^.    Therefore  a  straight      ^ 

line  has  been  erected  at  right  angles  to  a  given  plane  from  a 
point  given  in  it.     Which  was  to  be  done. 


a  11.  11. 
b  31.  1. 


o  8.  lU 


PROP.  XHL    THEOR. 


FROM  the  same  point  in  a  given  plane,  there  can- 
not  be  two  straight  lines  at  right  angles  to  the  plane, 
upon  the  same  side  of  it ;  and  there  can  be  but  one 
perpendicular  to  a  plane  from  a  point  above  the  plane. 

For,  if  it  be  possible,  let  the  two  straight  lines  AC,  AB  be  at 
right  angles  to  a  given  plane  from  the  same  point  A  in  the  plane, 
and  upon  the  same  side  of  it ;  and  let  a  plane  pass  through  BA, 


208 


THE  ELEMENTS 


Book  XI.  AC  ;  the  common  section  of  this  with  the  given  plane  is  a  straight* 
v.-^^^  line  passing  through  A  :  let  DAE  be  their  common  section :  there- 
a  3. 11.    fore  the  straight  lines  AB,  AC,  DAE  are  in  one  plane  :  and  be- 
cause CA  is  at  right  angles  to  the  given  plane,  it  shall  make 

right  angles  with  every  straight  line  meeting  it  in  that  plane. 

But  DAE,  which  is  in  that  plane,  meets  CA;  therefore  CAE  is 

a  right  angle.  For  the  same  reason 

BAE  is  a  right  angle.     Wherefore  B  C      " 

the  angle  CAE  is  equal  to  the  angle 

BAE;    and  they  are  in  one  plane, 

which  is  impossible.     Also,  from  a 

point  above  a  plane,  there  can  be 

but  one  perpendicular  to  that  plane  ; 

for,  if  there  could  be  two,  they  would  &  v 

b6. 11.    be  parallel  ^  to  one  another,  which      ^  A  li. 

is  absurd.     Therefore,  from  the  same  point,  &c.     Q.  E.  D. 


PROP.  XIV.    THEOR. 


PLANES  to  which  the  same  straight  line  is  per- 
pendicular, are  parallel  to  one  another. 


a  3.  def. 
11. 


b  17. 1. 


c  8.  def. 
11. 


Let  the  straight  line  AB  be  perpendicular  to  each  of  the  planes 
CD,  EF ;  these  planes  are  parallel  to  one  another. 

If  not,  they  shall  meet  one  another  when  produced  ;  let  them 
meet ;  their  common  section  shall  be  a 
straight  line  GH,  in  which  take  any 
point  K,  and  join  AK,  BK  :  then,  be- 
cause AB  is  perpendicular  to  the  plane 
EF,  it  is  perpendicular  »  to  the  straight 
line  BK  which  is  in  that  plane.  There- 
fore ABK  is  a  right  angle.  For  the 
same  reason,  BAK  is  a  right  angle  ; 
wherefore  the  two  angles  ABK,  BAK 
of  the  triangle  ABK  are  equal  to  two 
right  angles,  which  is  impossible  *» : 
therefore  the  planes  CD,  EF,  though 
produced,  do  not  meet  one  another ; 
that  is,  they  are  parallel  ^.  Therefore, 
planes,  &c.     Q.  E.  D. 


OF  EUCLIDu 


PROP.  XV.    THEOR. 

IF  two  straight  lines  meeting  one  another,  be  pa-  See  N. 
rallel  to  two  straight  lines  which  meet  one  another, 
but  are  not  in  the  same  plane  with  the  first  two ;  the 
plane  which  passes  through  these  is  parallel  to  the 
plane  passing  through  the  others. 

Let  AB,  BC,  two  straight  lines  meeting  one  another,  be  pa- 
rallel to  DE,  EF,  that  meet  one  another,  but  are  not  in  the  same 
plane  with  AB,  BC :  the  planes  through  AB,  BC,  and  DE,  EF 
shall  not  meet,  though  produced. 

From  the  point    B   draw    BG   perpendicular »  to  the    plane  a  11.  11. 
which  passes  through  DE,  EF,  and  let  it  meet  that  plane  in 
G ;   and  through   G  draw   GH    parallel  »>  to  ED,  and  GK  pa-  b  31. 1. 
rallel  to  EF  :  and  because   BG  is  perpendicular  to  the  plane 
through    DE,    EF,    it    shall  E 

make  right  angles  with  every 
straight  line  meeting  it  in  that 
plane  *=.  But  the  straight  lines 
GH,  GK  in  that  plane   meet 
it :  therefore  each  of  the  an- 
gles BGH,  BGK  is  a  right  an- 
gle :  and   because  BA  is  pa- 
rallel ^  to    GH   (for  each    of 
them  is  parallel  to  DE,  and 
they  are  not  both   in  the   same  plane  with  it)  the  angles  GBA, 
BGH  are  together  equal*  to  two  right  angles:  and  BGH  is  a  c  29. 1. 
right  angle;   therefore  also  GBA  is  a  right  angle,  and  GB  per- 
pendicular to  BA  :  for  the  same  reason,  GB  is  perpendicular  to 
BC :  since  therefore  the  straight  hne  GB  stands  at  right  angles 
to  the  two  straight  lines  BA,  BC,  that  cut  one  another  in  B, 
Gfi  is  perpendicular  f  to  the  plane  through  BA,  BC  ;  and  it  is  f  4. 11. 
perpendicular  to  the  plane  through  DE,  EF  :  therefore  BG  is 
perpendicular  to  each  of  the  planes  through  AB,  BC,  and  DE, 
EF  :  but  planes  to  which  the  same  straight  line  is  perpendicular, 
are  parallel  e  to  one  another  :  therefore  the  plane   through   AB,  g  14. 11. 
BC  is  parallel  to  the  plane  through  DE,J£F.    Wherefore,  if  two 
straight  lines,  Sec.    Q.  E.  D. 


(T^ 

F 

K 

\ 

C 

^^ 

c  3.  def. 
11. 

^...^^ 

D 

d  9. 11.. 

3D 


210 
Book  XI. 


THE  ELEMENTS 


SceN. 


PROP.  XVI.    THEOR. 

IF  two  parallel  planes  be  cut  by  another  plane,  their 
common  sections  with  it  are  parallels. 

Let  the  parallel  planes  AB,  CD  be  cut  by  the  plane  EFHG, 
and  let  their  common  sections  with  it  be  EF,  GH  :  EF  is  paral- 
lel to  GH. 

For,  if  it  is  not,  EF,  GH  shall  meet,  if  produced,  either  on 
the  side  of  FH,  or  EG  :  first,  let  them  be  produced  on  the  side 
of  FH,  and  meet  in  the  point  K  ;  therefore,  since  EFK  is  in 
the  plane  AB,  every  point  in  K 

EFK  is  in  that  plane  :  and  K 
is  a  point  in  EFK  ;  therefore 
K  is  in  the  plane  AB  :  for 
the  same  reason  K  is  also  in 
the  plane  CD :  wherefore  the 

planes     AB,     CD     produced  ^^  "*1  D 

meet  one  another  ;  but  they 
do  not  meet  since  they  are 
parallel  by  the  hypothesis : 
therefore  tke  straight  lines 
EF,  GH  do  not  meet  when 
produced  on  the  side  of  FH ;  in  the  same  manner  it  may  be 
.proved,  that  EF,  GH  do  not  meet  when  produced  on  the  side 
of  EG  :  but  straight  lines  which  are  in  the  same  plane  and  do 
not  meet,  though  produced  either  way,  are  parallel :  therefore 
EF  is  parallel  to  GH.  Wherefore,  if  two  parallel  planes,  &c. 
Q.  E.  D. 


h' 

/ 

fA 

^ 

B 

k. 

C 

L^ 

E 

'^ 

G 

H 


PROP.  XVH.    THEOR. 

IF  two  Straight  lines  be  cut  by  parallel  planes,  thev 
shall  be  cut  in  the  same  ratio. 

Let  the  straight  lines  AB,  CD  be  cut  by  the  parallel  planes 
GH,  KL,  MN,  in  the  points  A,  E,  B ;  C,  F,  D  :  as  AE  is  to  EB, 
so  is  CF  to  FD. 

Join  AC,  BD,  AD,  and  let  AD  meet  the  plane  KL  in  the 
point  X ;  and  join  EX,  XF :  because  the  two  parallel  planes 
KL,  MN  are  cut  by  the  plane  EBDX,  the    common    sections 


OF  EUCLID. 


$11 


b  2.  6. 


EX,  Bl)  are  parallel".    For  the  same  reasorii  because  the  two  Book  XI. 
parallel  planes  GH,  KL  are  cut  ^-v^ 

by  the  plane  AXFC,  the  com-  /  _- — "!>.         /  H  a  16. 11. 

men  sections  AC,  XF  are  paral- 
lel: and  because  EX  is  paral- 
lel to  BD,  a  side  of  the  triangle 
ABD,  as  AE  to  EB  so  is  ^  AX 
to  XD.  Again,  because  XF  is 
parallel  to  AC,  a  side  of  the 
triangle  ADC,  as  AX  to  XD, 
so  is  CF  to  FD :  and  it  was 
proved  that  AX  is  to  XD  as 
AE  to  EB :  therefore  <=,  as 
AE  to  EB  so  is  CF  to  FD. 
Wherefore,  if  two  straight  lines, 
kc,    Q.  E.  D. 


c  11.  S; 


PROP.  XVIII.     THEOR. 


IF  a  straight  line  be  at  right  angles  to  a  plane,  eve- 
ry plane  which  passes  through  it  shall  be  at  right  an- 
gles to  that  plane. 

Let  the  straight  line  AB  be  at  right  angles  to  a  plane  CK ; 
every  plane  which  passes  through  AB  shall  be  at  right  angles  to, 
the  plane  CK. 

Let  any  plane  DE  pass  through  AB,  and  let  CE  be  the  com- 
mon section  of  the  planes  DE,  CK ;  take  any  point  F  in  CE, 
from  which  draw  FG  in  the 
plane  DE  at  right  angles  to 
CE :  and  because  AB  is  per- 
pendicular to  the  plane  CK, 
therefore  it  is  also  perpendi- 
cular to  every  straight  line  in 
that  plane  meeting  it  »  ;  and 
consequently  it  is  perpendicu- 
lar to  CE  :  wherefore  ABF 
is  a  right  angle ;  but  GFB  is 
likewise  a  right  angle:   there- 


D             G 

A         H 

' ' 

K 

\ 

\ 

a  3.  de£. 
11. 


B 


E 


fore  AB  is  parallel  ^  to  FG.     And  AB  is  at  right  angles  to  the  b  28.  t 
plane  CK ;    therefore  FG  is  also  at  right  angles  to  the  same 
plane  «.     But  one  plane  is  at  right  angles  to  another  plane  when  c  8. 11. 
the  straight  lines  drawn  in  one  of  the  planes,  at  right  angles  to 
their  common  section,   are  also  at  right  angles  to  the  other 


212 


THE  ELEMENTS 


Book  XI.  plane  d :    and  any  straight  line  FG  in  the  plane  DC,  which  is 

y^'-Y'mJ  at  right  angles  to  CE  the  common  section  of  the  planes,   has 

d4.  def.    been  proved  to  be  perpendicular  to  the  other  plane  CK ;    there- 

II-        fore  the  plane  DE  is  at  rigist  angles  to  the  plane  CK.     In  like 

manner  it  may  be  proved  that  all  the  planes  which  pass  through 

AB  are  at  right  angles  to  the  plane  CK.    Therefore,  if  a  straight 

line,  £cc.    Q.  E.  D. 


PROP.  XIX.    THEOR. 

IF  two  planes  cutting  one  another  be  each  of  them 
perpendicular  to  a  third  plane,  their  common  section 
shall  be  perpendicular  to  the  same  plane. 


a  4  def. 
11. 


b  13. 11 


Let  the  two  planes  AB,  BC  be  each  of  them  perpendicular  to 
a  third  plane,  and  let  BD  be  the  common  section  of  the  first  two ; 
BD  is  perpendicular  to  the  third  plane. 

If  it  be  not,  from  the  point  D  draw,  in  the  plane  AB,  the 
straight  line  DE  at  right  angles  to  AD  the  common  section  of 
the  plane  AB  with  the  third  plane ;  and  in  the  plane  BC  draw 
DF  at  right  angles  to  CD  the  common  section  of  the  plane  BC 
with  the  third  plane.  And  because  the 
plane  AB  is  perpendicular  to  the  third 
plane,  and  DE  is  drawn  in  the  plane  AB 
at  right  angles  to  AD  their  common  sec- 
tion, DE  is  perpendicular  to  the  third 
plane  a.  In  the  same  manner  it  may  be 
proved  that  DF  is  perpendicular  to  the 
third  plane.  Wherefore,  from  the  point 
D  two  straight  lines  stand  at  right  angles 
to  the  third  plane,  upon  the  same  side  of 
it,  which  is  impossible^:  therefore  from 
the  point  D  there  cannot  be  any  straight 
line  at  right  angles  to  the  third  plane, 
except  BD  the  common  section  of  the       ^  ^ 

planes  AB,   BC      BD  therefore   is  perpendicular  to  the   third 
plane.     Wherefore,  if  two  planes,  Sec.     Q.  E.  D. 


OF  EUCLID. 


PROP.  XX.    THEOR. 


IF  a  solid  angle  be  contained  by  three  plane  angles,  See  N. 
any  two  of  them  are  greater  than  the  third. 

Let  the  solid  angle  at  A  be  contained  by  the  three  plane  an- 
gles BAG,  CAD,  DAB.  Any  two  of  them  are  greater  than  the 
third. 

If  the  angles  BAC,  CAD,  DAB  be  all  equal,  it  is  evident  that 
any  two  of  them  are  greater  than  the  third.  But  if  they  are  not, 
let  BAC  be  that  angle  which  is  not  less  than  either  of  the  other 
two,  and  is  greater  than  one  of  them  DAB ;  and  at  the  point  A, 
in  the  straight  line  AB,  make,  in  the  plane  which  passes  through 
BA,  AC,  the  angle  BAE  equal  »  to  the  angle  DAB  ;  and  make  a  23.1. 
AE  equal  to  AD,  and  through  E  draw  BEC  cutting  AB,  AC  in 
the  points  B,  C,  and  join  DB,  DC.  And  because  DA  is  equal  to 
AE,  and  AB  is  common,  the  two  DA, 
AB  are  equal  to  the  two  EA,  AB,  and  D 

the  angle  DAB  is  equal    to  the  angle  /k 

EAB  :  therefore  the  base  DB  is  equal''  /  1     ><  b  4, 1. 

to  the  base  BE.     And  because  BD,  DC  /I       \^ 

are  greater  c  than  CB,  and  one  of  them        /  ^  |  \^  ^  ^"*  *' 

BD  has  been  proved  equal  to  BE  a  part 
of  CB,  therefore  the  other  DC  is  great-  ^^ 
er  than  the  remaining  part  EC.  And  g 
^  because  DA  is  equal  to  AE,  and  AC 
common,  but  the  base  DC  greater  than  the  base  EC  :  therefore 
the  angle  D AC  is  greater  ^  than  the  angle  EAC ;  and,  by  the  d  25. 1. 
construction,  the  angle  DAB  is  equal  to  the  angle  BAE  ;  where- 
fore the  angles  DAB,  DAC  are  together  greater  than  BAE,  EAC, 
that  is,  than  the  angle  BAC.  But  BAC  is  not  less  than  either 
of  the  angles  DAB,  DAC  :  therefore  BAC,  with  either  of  them, 
is  greater  than  the  other.  Wherefore,  if  a  solid  angle,  8cc. 
Q.  E.  D. 

PROP.  XXI.    THEOR. 

EVERY  solid  angle  is  contained  by  plane  angles 
which  together  are  less  than  four  right  angles. 

First,  let  the  solid  angle  at  A  be  contained  by  three  plane  angles 
BAC,  CAD,  DAB.  These  three  together  are  less  than  four 
right  angles. 


21* 


THE  ELEMENTS 


Book  XI.      Take  in  each  of  the  straight  lines  AB,  AC,  AD  any  points  B, 

*«— v^"-*  C,  D,  and  join  BC,  CD,  DB  :  then  because  the  solid  angle  at  B 
is  contained  by  the  three  plane  angles  CBA,   ABD,  DBC,  any 

a  20. 11.  two  of  them  are  greater  »  than  the  third  ;  therefore  the  angles 
CBA,  ABD  are  greater  than  the  angle  DBC  :  for  the  same  rea- 
son, the  angles  BCA,  ACD  are  greater  than  the  angle  DCB ; 
and  the  angles  CDA,  ADB  greater  than  BDC  :  wherefore  the 
six  angles  CBA,  ABD,  BCA,  ACD,  CDA,  ADB  are  greater  than 
the  three  angles  DBC,  BCD,  CDB  :  but 
the  three  angles  DBC,  BCD,  CDB  are 

b32.  1.  equal  to  two  right  angles  *>  :  therefore 
the  six  angles  CBA,  ABD,  BCA,  ACD, 
CDA,  ADB  are  greater  than  two  right 
angles  :  and  because  the  three  angles 
of  each  of  the  triangles  ABC,  ACD, 
ADB  are  equal  to  two  right  angles, 
therefore  the  nine  angles  of  these  three  triangles,  viz.  the  angles 
CBA,  BAC,  ACB,  ACD,  CDA,  DAC,  ADB,  DBA,  BAD  are 
equal  to  six  right  angles  :  of  these  the  six  angles  CBA,  ACB, 
ACD,  CDA,  ADB,  DBA  are  greater  than  two  right  angles  : 
therefore  the  remaining  three  angles  BAC,  DAC,  BAD,  which 
contain  the  solid  angle  at  A,  are  less  than  four  right  angles. 

Next,  let  the  solid  angle  at  A  be  contained  by  any  number  of 
plane  angles  BAC,  CAD,  DAE,  EAF,  FAB  ;  these  together  are 
less  than  four  right  angles. 

Let  the  planes  in  which  the  angles  are  be  cut  by  a  plane,  and 
let  the  common  section  of  it  Avith  those 
planes  be  BC,  CD,  DE,  EF,  FB  :  and 
because  the  solid  angle  at  B  is  contain- 
ed by  three  plane  angles  CBA,  ABF, 
FBC,  of  which  any  two  are  greater » 
than  the  third,  the  angles  CBA,  ABF 
are  greater  than  the  angle  FBC:  for 
the  same  reason,  the  two  plane  angles 
at  each  of  the  points  C,  D,  E,  F,  viz.  the 
angles  which  are  at  the  bases  of  the 
triangles,  having  the  common  vertex 
A,  are  greater  than  the  third  angle  at 
the  same  point,  which  is  one  of  the  an- 
gles of  the  polygon  BCDEF :  there- 
fore all  the  angles  at  the  bases  of  the  triangles  are  together 


OF  EUCLID.  ^]s 

greater  than  all  the  angles  of  the  polygon  :  and  because  all  the  Book  XL 
angles  of  the  triangles  are  together  equal  to  twice  as  many  right  v  -y^ 
angles  as  there  are  triangles  ^  ;  that  is,  as  there  arc  sides  in  the  b  32. 1. 
polygon  BCDEF  :  and  that  all  the  angles  of  the  polygon,  toge- 
ther with  four  right  angles,  are  likewise  equal  to  twice  as  many 
right  angles  as  there  are  sides  in  the  polygon  <=  ;  therefore  all  c  1.  Coi;^ 
the  angles  of  the  triangles  are  equal  to  all  the  angles  of  the  po-   32. 1. 
lygon  together  with  four  right  angles.     But  all  the  angles  at  the 
bases  of  the  triangles  are  greater  than  all  the  angles  of  the  poly- 
gon, as  has  been  proved.     Wherefore  the  remaining  angles  of 
the  triangles,  viz.  those  at  the  vertex,  which  contain  the  solid 
angle  at  A,  are  less  than  four  right  angles.     Therefore,  every 
solid  angle,  &c.     Q.  £.  D. 


PROP.  XXII.    THEOH. 


IF  every  two  of  three  plane  angles  be  greater  than  see  ^^ 
the  third,  and  if  the  straight  lines  which  contain  them 
be  all  equal ;  a  triangle  may  be  made  of  the  straight 
lines  that  join  the  extremities  of  those  equal  straight 
lines. 


Let  ABC,  DEF,  GHK  be  three  plane  angles,  whereof  eveiy 
two  are  greater  than  the  third,  and  are  contained  by  the  equal 
straight  lines  AB,  BC,  DE,  EF,  GH,  HK ;  if  their  extremities 
be  joined  by  the  straight  lines  AC,  DF,  GK,  a  triangle  may  be 
made  of  three  straight  lines  equal  to  AC,  DF,  GK  ;  that  is,  eve- 
ry two  of  them  are  together  greater  than  the  third. 

If  the  angles  at  B,  E,  H  are  equal,   AC,  DF,  GK  are  also 
equal  a,   and  any  two  of  them   greater  than  the  third:   but  if  a  41. 
the  angles  are  not  all  equal,  let  the  angle  ABC  be  not  less  than 
cither  of  the   two  at  E,    H;    therefore  the   straight  line   AC 
is  not  less  than  either  of  the  other  two  DF,  GK  '> ;  and  it  is  b  4.  or 
plain  that  AC,  together  with  either  of  the  other  two,  must  be  24. 1. 
greater  than  the  third  :    also  DF  with  GK  are  greater  than 
AC:    for,    at  the  point  B  in  the  straight  line  AB  make  <^  the  c  23. 1. 


316 


THE  ELEMENTS 


Book XI.  angle  ABL  equal  to  the  an^le  GHK,  and  make  BL  equal  to 
**— V— ^  one  of  the  straight  lines  AB,  BC,  DE,  EF,  GH,  HK,  and  join 
AL,  LC  ;  then,  because  AB,  BL  are  equal  to  GH,  HK,  and  the 
angle  ABL  to  the  angle  GHK,  the  base  AL  is  equal  to  the  base 
GK  :  and  because  the  angles  at  E,  H  are  greater  than  the  angle 
ABC,  of  which  the  angle  at  H  is  equal  to  ABL  ;  therefore  the 
remaining  angle  at  E  is  greater  than  the  angle  LBC  :  and  be- 


F      G 


d  24. 1. 


e20. 1. 


£23. 1. 


cause  the  two  sides  LB,  BC  are  equal  to  the  two  DE,  EF,  and 
that  the  angle  DEF  is  greater  than  the  angle  LBC,  the  base  DF 
is  greater  ^  than  the  base  LC :  and  it  has  been  proved  that  GK 
is  equal  to  AL ;  therefore  DF  and  GK  are  greater  than  AL  and 
LC  :  but  AL  and  LC  are  greater  ^  than  AC :  much  more  then 
are  DF  and  GK  greater  than  AC.  Wherefore  every  two  of 
these  straight  lines  AC,  DF,  GK  are  greater  than  the  third  ;  and, 
therefore,  a  triangle  may  be  made  f,  the  sides  of  which  shall  be, 
equal  to  AC,  DF,  GK.    Q.  E.  D. 


PROP.  XXHL    PROB. 


See  N  TO  make  a  solid  angle  which  shall  be  contained 
by  three  given  plane  angles,  any  two  of  them  being 
greater  than  the  third,  and  all  three  together  less  than 
four  right  angles. 

Let  the  three  given  plane  angles  be  ABC,  DEF,  GHK,  any 
two  of  which  are  greater  than  the  third,  and  all  of  them  toge- 
ther less  than  four  right  angles.  It  is  required  to  make  a  solid 
angle  contained  by  three  plane  angles  equal  to  ABC,  DEF, 
GHK,  each  to  each. 


OF  EUCLID. 


217 


From  the  straight  lines  containing  the  angles,  cut  off  AB,  BC,  BookXI. 
DE,  EF,  GH,  HK,  all  equal  to  one  another;  and  join  AC,  DF,  ' — r— ' 
GK :  then  a  triangle  may  be  made  »  of  three  straight  lines  ec^ual  a  22. 11. 


to  AC,  DF,  GK.     Let  this  be  the  triangle  LMN^,  so  that  AC  b  22.  t 
be  equal  to  LM,  DF  to  MN,  and  GK  to  LN  ;  and  about  the  tri- 
angle LMN  describe  *=  a  circle,  and  find  its  centre  X,  which  will  c  5. 4' 
cither  be  within  the  triangle,  or  in  one  of  its  sides,  or  without  it. 
First,  let  the  centre  X  be  within  the  triangle,  and  join  LX, 
MX,  NX  :  AB  is  greater  than  LX :  if  not,  AB  must  <nther  be 
equal  to,  or  less  than  LX ;  first,  let  it  be  equal:  then  because 
AB  is  equal  to  LX,  and  that  AB  is  also  equal  to  BC,  and  LX  to 
XM,  AB  and  BC  are  equal  to  LX  and  XM,  each  to  each  ;  and 
the  base  AC  is,  bv  construction,  equal  to  the  base  LM :  where- 
fore the  angle  ABC  is  equal  to  the  angle  LXM^.     For  the  same  d  8  1. 
reason,  the  angle  DEF  is  equal  to  the  angle  MXN,  and  the  an- 
gle  GHK   to    the    angle   NXL; 
therefore  the   three  angles  ABC, 
DEF,  GHK  are  equal  to  the  three 
angles  LXM,  MXN,  NXL :  but  the 
three  angles  LXM,  MXJ^,  NXL 
are  equal  to  four  right  angles  <= : 
therefore    also   the    three    angles 
ABC,  DEF,  GHK  are  equal  to  four 
right  angles  :  but,  by  the  hypothe- 
sis, they  are  less  than  four  right  an- 
gles ;  which  is  absurd ;  therefore 
AB  is  not  equal  to  LX :  but  nei- 
ther can   AB   be    less  than   LX : 
for,  if  possible,  let  it  be  less,  and 
upon    the    straight   line    LM,    on 
the  side  of  it  on  which  is  the  cen- 
tre X,  describe  the  triangle  LOM, 

the  sides  LO,  OM  of  which  are  equal  to  AB,  BC ;  and  because 
the  base  LM  is  equal  to  the  base  AC,  the  angle  LOM  is  equal 

2  E 


e  2.  Cor. 
15. 1. 


218 


THE  ELEMENTS 


Book  XI.  to  the  angle  ABC^^:  and  AB,  that  is,  LO,  by  the  hypothesis,  is 

V — y^  less  than  LX  ;  wherefore  LO,  OM  fall  within  the  triangle  LXM  ; 

d  8. 1.  for,  if  they  fell  upon  its  sides,  or  without  it,  they  would  be  equal 
to,  or  greater  than  LX,  XM^:  therefore  the  angle  LOM,  that 
is,  the  angle  ABC,  is  greater  than 

f  21. 1.  the  angle  LXM  f :  in  the  same  man- 
ner it  may  be  proved  that  the  angle 
DEF  is  greater  than  the  angle  MXN, 
and  the  angle  GHK  greater  than  the 
angle  NXL.  Therefore  the  three 
angles  ABC,  DEF,  GHK  are  great- 
er than  the  three  angles  LXM,  MXN, 
NXL ;  that  is,  than  four  right  an- 
gles :  but  the  same  angles  ABC, 
DEF,  GHK  are  less  than  four  right 
angles;  which  is  absurd:  therefore 
AB  is  not  less  than  LX,  and  it  has 
been  proved  that  it  is  not  equal  to 
LX ;  wherefore  AB  is  greater  than 
LX. 

Next,  let  the  centre  X  of  the  circle  fall  in  one  of  the  sides  of 
the  triangle,  viz.  in  MN,  and  join 
XL :  in  this  case  also  AB  is  great- 
er than  LX.  If  not,  AB  is  either 
equal  to  LX,  or  less  than  it :  first, 
let  it  be  equal  to  XL  :  therefore  AB 
and  BC,  that  is,  DE,  and  EF,  are 
equal  to  MX  and  XL,  that  is,  to 
MN :  but,  by  the  construction, 
MN  is  equal  to  DF  ;  therefore  DE, 
EF  are  equal  to  DF,  which  is  im- 

t  20. 1.  possible  t :  wherefore  AB  is  not 
equal  to  LX  :  nor  is  it  less  :  for 
then,  much  more,  an  absurdity 
would  follow :  therefore  AB  is 
greater  than  LX. 

But,  let  the  centre  X  of  the  circle  fall  without  the  triangle 
LMN,  and  join  LX,  MX,  NX.  In  this  case  likewise  AB  is 
greater  than  LX  :  if  not,  it  is  either  equal  to,  or  less  than  LX  : 
first,  let  it  be  equal ;  it  may  be  proved  in  the  same  manner, 
as  in  the  first  case,  that  the  angle  ABC  is  equal  to  the  angle 
MXL,  and  GHK  to  LXN;  therefore  the  whole  angle  MXN 
is  equal  to  the  two  angles  ABC,  GHK  :  but  ABC  and  GHK 
are  together  greater  than  the  angle  DEF  ;  therefore  also 
the   angle   MXN   is   greater   thao    DEF.      And  because    DE. 


OF  EUCLID. 


219 


EF  are  equal  to  MX,  XN,  and  the  base  DF  to  the  base  MN,  the  Book  XI. 
angle  MXNT  is  equal  ^  to  the  angle  DEF :   and  it  has  been  prov-  ^-v— •> 
ed  that  it  is  greater  than  DEF,  which  is  absurd.    Therefore  AB  d  8. 1. 
is  not  equal  to  LX.      Nor  yet  is  it  less;    for  then,  as  has  been 
proved  in  the  first  case,  the  angle  ABC  is  greater  than  the  angle 
MXL,  and  the  angle  GHK  greater  than  the  angle  LXN.      At 
the  point  B,  in  the  straight  line  CB,  make  the  angle  CBP  equal 
to  the  angle  GHK,  and  make  BP  equal  to  HK,  and  join  CP,  AP. 


And  because  CB  is  equal  to  GH ;  CB,  BP  are  equal  to  GH,  HK, 
each  to  each,  and  they  contain  equal  angles ;  wherefore  the 
base  CP  is  equal  to  the  base  GK,  that  is,  to  LN.  And  in  the 
isosceles  triangles  ABC,  MXL,  because  the  angle  ABC  is  great- 
er than  the  angle  MXL,  therefore  the  angle  MLX  at  the  base  is 
greater  s  than  the  angle  ACB  at  the  base.  For  the  same  rea-  g  32.  >. 
son,  because  the  angle  GHK,  or 

CBP,  is  greater  than  the  angle  R 

LXN,  the  angle  XLN  is  greater 
than  the  angle  BCP.  Therefore 
the  whole  angle  MLN  is  greater 
than  the  whole  angle  ACP.  And 
because  ML,  LN  are  equal  to 
AC,  CP,  each  to  each,  but  the 
angle  MLN  is  greater  than  the 
angle  ACP,  the  base  MN  is 
greater  ^  than  the  base  AP. 
And  MN  is  equal  to  DF  ;  there- 
fore also  DF  is  greater  than  AP. 
Again,  because  DE,EF  are  equal 
to  AB,  BP,  but  the  base  DF 
greater  than  the  base  AP,  the 
angle  DEF  is  greater  ^  than  the  1-  35.  l 

angle  ABP.  And  ABP  is  equal  to  the  two  angles  ABC,  CBP, 
that  is,  to  the  two  angles  ABC,  GHK ;  therefore  the  angle  DEF 
is  greater  than  the  two  angles  ABC,  GHK  ;   but  it  is  also  less 


h  24. 1. 


220 


THE  ELEMENTS 


b  3.  def. 
11. 


Book  XI.  than  these,  which  is  impossible.     Therefore  AB  is  not  less  than 

Ci-y-.^  LX,  and  it  has  been  proved  that  it  is  not  equal  to  it ;  therefore 
AB  is  greater  than  LX. 

a  12. 11.  From  the  point  X  erect  a  XR  at  right  angles  to  the  plane  of 
the  circle  LMN.  And  because  it  has  been  proved  in  all  the  cases 
that  AB  is  greater  than  LX,  find  a  square  equal  to  the  excess  of 
the  square  of  AB  above  the  square 
of  LX;  and  make  RX  equal  to  its 
side,  and  join  RL,  RM,  RN.  Be- 
cause RX  is  perpendicular  to  the 
plane  of  the  circle  LMN,  it  is  •>  per- 
pendicular to  each  of  the  straight 
lines  LX,  MX,  NX.  And  because 
LX  is  equal  to  MX,  and  XR  com- 
mon, and  at  right  angles  to  each  of 
them,  the  base  RL  is  equal  to  the 
base  RM.  For  the  same  reason, 
RN  is  equal  to  each  of  the  two  RL, 
RM.  Therefore  the  three  straight 
lines  RL,  RM,  RN  are  all  equal. 
And  because  the  square  of  XR  is 
equal  to  the  excess  of  the  square 
of  AB  above  the  square  of  LX ; 
therefore  the  square  of  AB  is  equal  to  the  squares  of  LX,  XR. 

c4r.  1.  But  the  square  of  RL  is  equal  <=  to  the  same  squares,  because 
LXR  is  a  right  angle.  Therefore  the  square  of  AB  is  equal  to 
the  square  of  RL,  and  the  straight  line  AB  to  RL.  But  each  of 
the  straight  lines  BC,  DE,  EF,  GH,  HK  is  equal  to  AB,  and 
each  of  the  two  RM,  RN  is  equal  to  RL.  Wherefore  AB,  BC, 
DE,  EF,  GH,  HK  are  each  of  them  equal  to  each  of  the  straight 
lines  RL,  RM,  RN.  And  because  RL,  RM  are  equal  to  AB, 
BC,  and  the  base  LM  to  the  base  AC ;  the  angle  LRM  is  equal 

d  8. 1.  ^  to  the  angle  ABC.  For  the  same  reason,  the  angle  MRN  is 
equal  to  the  angle  DEF,  and  NRL  to  GHK.  Therefore  there  is 
made  a  solid  angle  at  R,  which  is  contained  by  three  plane  an- 
gles LRM,  MRN,  NRL,  which  are  equal  to  the  three  given 
plane  angles  ABC,  DEF,  GHK,  each  to  each.  Which  was  to 
be  done. 


OF  EUCLID* 


PROP.  A.    THEOR. 


IF  each  of  two  solid  angles  be  contained  by  three  see  n. 
plane  angles  equal  to  one  another,  each  to  each;  the 
planes  in  which  the  equal  angles  are,  have  the  same 
inclination  to  one  another. 


a  6. 
def.  U 


Let  there  be  two  solid  angles  at  the  points  A,  B ;  and  let 
the  angle  at  A  be  contained  by  the  three  plane  angles  CAD, 
CAE,  EAD  ;  and  the  angle  at  B  by  the  three  plane  angles 
FBG,  FBH,  HBG;  of  which  the  angle  CAD  is  equal  to  the 
angle  FBG,  and  CAE  to  FBH,  and  EAD  to  HBG  :  the  planes  in 
which  the  equal  angles  are  have  the  same  inclination  to  one 
another. 

In  the  straight  line  AC  take  any  point  K,  and  in  the  plane 
CAD  from  K  draw  the  straight  line  KD  at  right  angles 
to  AC,  and  in  the 
plane  CAE  the  straight 
line  KL  at  right  angles 
to  the  same  AC : 
therefore  the  angle 
DKL  is  the  inclina- 
tion a  of  the  plane 
CAD  to  the  plane 
CAE.  In  BF  take 
BM  equal  to  AK,  and 
from  the  point  M  draw,  in  the  planes  FBG,  FBH,  the  straight 
lines  MG,  MN  at  right  angles  to  BF ;  therefore  the  angle  GMN 
is  the  inclination  a  of  the  plane  FBG  to  the  plane  FBH :  join 
LD,  NG;  and  because  in  the  triangles  KAD,  MBG,  the 
angles  KAD,  MBG  are  equal,  as  also  the  right  angles  AKD, 
BMG,  and  that  the  sides  AK,  BM,  adjacent  to  the  equal 
angles,  are  equal  to  one  another;  therefore  KD  is  equal  ^  to  b  26.  1 
MG,  and  AD  to  BG :  for  the  same  reason,  in  the  triangles 
KAL,  MBN,  KL  is  equal  to  MN,  and  AL  to  BN :  and  in  the 
triangles  LAD,  NBG,  LA,  AD  are  equal  to  NB,  BG,  and  they 
contam  equal  angles;  therefore  the  base  LD  is  equal  <=  to  the  c  4. 1 
base  NG.  Lastly,  in  the  triangles  KLD,  MNG,  the  sides  DK, 
KL  are  equal  to  GM,  MN,  and  the  base  LD  to  the  base  NG; 
therefore  the  angle  DKL  is  equal  d  to  the  angle  GMN  :  but  the  d  8. 1. 
angle  DKL  is  the  inclination  of  the  plane  CAD  to  the  plane 
CAE,  and  the  angle  GMN  is  the  inclination  of  the  plane  FBG 


222  THE  ELEMENTS 

Book  XI.  to  the  plane  FBH,  which  planes  have  therefore  the  same  incli- 
*— V— ^  nation  »  to  one  another :  and  in  the  same  manner  it  may  be  de- 
a  7.  monstrated,  that  the  other  planes  in  which  the  equal  angles  are 

dof.  11.    Yiave  the  same  inclination  to  one  another.     Therefore,  if  two  so- 
lid angles,  See.     Q.  E.  D. 


PROP.  B.    THEOR. 

See  N,  IF  two  solid  angles  be  contained,  each  by  three 
plane  angles  which  are  equal  to  one  another,  each  to 
each,  and  alike  situated;  these  solid  angles  are  equal 
to  one  another. 

Let  there  be  two  solid  angles  at  A  and  B,  of  which  the  solid 
angle  at  A  is  contained  by  the  three  plane  angles  CAD,  CAE, 
EAD ;  and  that  at  B  by  the  three  plane  angles  FBG,  FBH, 
HBG  ;  of  which  CAD  is  equal  to  FBG ;  CAE  to  FBH ;  and 
EAD  to  HBG  :  the  solid  angle  at  A  is  equal  to  the  solid  angle 
at  B. 

Let  the  solid  angle  at  A  be  applied  to  the  solid  angle  at  B ; 
and,  first,  the  plane  angle  CAD  being  appHed  to  the  plane  angle 
FBG,  so  as  the  point  A  may  coincide  with  the  point  B,  and  the 
straight  line  AC  with  BF ;  then  AD  coincides  wiih  BG,  because 
the  angle  CAD  is  equal  to  the 
angle  FBG  :  and  because  the  in- 
clination of  the  plane  CAE  to  the 

a  A.  11.  plane  CAD  is  equal »  to  the  in- 
clination of  the  plane  FBH  to  the 
plane  FBG,  the  plane  CAE  coin- 
cides with  the  plane  FBH,  be- 
cause the  planes  CAD,  FBG  co-  C 
incide  with  one  another :  and  because  the  straight  lines  AC,  BF 
coincide,  and  that  the  angle  CAE  is  equal  to  the  angle  FBH ; 
therefore  AE  coincides  with  BH,  and  AD  coincides  with  BG ; 
wherefore  the  plane  EAD  coincides  with  the  plane  HBG  :  there- 
fore the  scHd  angle  A  coincides  with  the  solid  angle  B,  and  con- 

b  8.  A.  1.  sequently  they  are  equal  *>  to  one  another.     Q.  E.  D. 


OF  EUCLID, 


PROP.  C.     THEOR. 


'    SOLID  figures  contained  by  the  same  number  of  See  n. 
equal  and  similar  planes  alike  situated,  and  having 
none  of  their  solid  angles  contained  by  more  than 
three  plane  angles,  are  equal  and  similar  to  one  an- 
other. 


Let  AG,  KQ  be  two  solid  figures  contained  by  the  same  num- 
ber of  similar  and  equal  planes,  alike  situated,  viz.  let  the  plane 
AC  be  similar  and  equal  to  the  plane  KM,  the  plane  AF  to  KP; 
BG  to  LQ  ;  GD  to  QN  ;  DE  to  NO  ;  and  lastly,  FH  similar  and 
equal  to  PR  :  the  solid  figure  AG  is  equal  and  similar  to  the  so- 
lid figure  KQ. 

Because  the  solid  angle  at  A  is  contained  by  the  three  plane 
angles  BAD,  BAE,  EAD,  which,  by  the  hj^pothesis,  are  equal 
to  the  plane  angles  LKN,  LKO,  OKN,  which  contain  the  solid 
angle  at  K,  each  to  each  ;  therefore  the  solid  angle  at  A  is  equal  ^  a  B.  11. 
to  the  solid  angle  at  K :  in  the  same  manner,  the  other  solid  an- 
gles of  the  figures  are  equal  to  one  another.  If,  then,,  the  solid 
figure  AG  be  applied  to  the  solid  figure  KQ,  first,  the  plane 
figure  AC  being  ap- 

H  G     . 


R 


\  E 

\ 

C 

\ 

\ 

O 


Q 


Mi 


A 


B 


K 


plied  to  the  plane 
figure  KM ;  the 
straight  line  AB  co- 
inciding with  KL, 
the  figure  AC  must  ^ 
coincide  with  the 
figure  KM,  because 
they  are  equal  and 
similar :  therefore  the  straight  lines  AD,  DC,  CB  coincide  with 
KN,  NM,  ML,  each  with  each  ;  and  the  points  A,  D,  C,  B,  with 
the  points  K,  N,  M,  L :  and  the  solid  angle  at  A  coincides  with* 
the  solid  angle  at  K ;  wherefore  the  plane  AF  coincides  with 
the  plane  KP,  and  the  figure  AF  with  the  figure  KP,  because 
they  are  equal  and  similar  to  one  another:  therefore  the  straight 
lines  AE,  EF,  FB  coincide  with  KO,  OP,  PL  ;  and  the  points 
E,  F  with  the  points  O,  P.  In  the  same  manner,  the  figure 
AH  coincides  with  the  figure  KR,  and  the  straight  line  DH  with 
NR,  and  the  point  H  with  the  point  R :  and  because  the  solid 
angle  at  B  is  equal  to  the  solid  angle  at  L,  it  may  be  proved,  in 
the  sanae  manner,  that  the  figure  BG  coincides  with  the  figure 


224 


THE  ELEMENTS 


Book  XL  LQ,  and  the  straight  line  CG  with  MQ,  and  the  point  G  with 
^-— v-i-^  the  point  Q:  since,  therefore,  all  the  planes  and  sides  of  the  so- 
lid figure  AG  coincide  with  the  planes  and  sides  of  the  solid  fi- 
gure KQ,  AG  is  equal  and  similar  to  KQ :  and,  in  the  same  man- 
ner, any  other  solid  figures  whatever  contained  by  the  same  num- 
ber of  equal  and  similar  planes,  alike  situated,  and  having  none 
of  their  solid  angles  contained  by  more  than  three  plane  angles, 
may  be  proved  to  be  equal  and  similar  to  one  another.    Q.  E.  D. 


PROP.  XXIV.    THEOR. 


SeeN. 


IF  a  solid  be  contained  by  six  planes,  two  and  two 
of  which  are  parallel ;  the  opposite  planes  are  similar 
and  equal  parallelograms. 


Let  the  solid  CDGH  be  contained  by  the  parallel  planes  AC, 
GF;  BG,  CE  ;  FB,  AE  :  its  opposite  planes  are  similar  and 
equal  parallelograms. 

Because  the  two  parallel  planes  BG,  CE  are  cut  by  the  plane 

a  16.  11.  AC,  their  common  sections  AB,  CD  are  parallel  ».  Again,  be- 
cause the  two  parallel  planes  BF,  AE  are  cut  by  the  plane  AC, 
their  common  sections  AD,  BC  are  parallel*:  and  AB  is  paral- 
lel to  CD  ;  therefore  AC  is  a  parallelogram.  In  like  manner,  it 
may  be  proved  that  each  of  the  figures 

CE,  FG,  GB,  BF,  AE  is  a  parallelo-  B  H 

gram  :  join  AH,  DF  ;  and  because 
AB  is  parallel  to  DC,  and  BH,  to  CF; 
the  two  straight  lines  AB,  BH  which  A 
meet  one  another,  are  parallel  to  DC 
and  CF  which  meet  one  another,  and 
are  not  in  the  same  plane  with  the 
other  two;    wherefore  they  contain 

b  10.  11.  equal   angles  ^  ;    the   angle   ABH  is      D 
therefore   equal  to  the  angle  DCF  : 
and  because  AB,  BH  are  equal  to  DC,  CF,  and  the  angle  ABH 

c  4. 1.  equal  to  the  angle  DCF  ;  therefore  the  base  AH  is  equal  <=  to  the 
base  DF,  and  the  triangle  ABH  to  the  triangle  DCF  :  and  the 

d34. 1.  parallelogram  BG  is  double  ^  of  the  triangle  ABH,  and  the  pa- 
rallelogram CE  double  of  the  triangle  DCF;  therefore  the  paral- 
lelogram BG  is  equal  and  similar  to  the  parallelogram  CE.  In 
the  same  manner  it  may  be  proved,  that  the  parallelogram  AC 


^^--^ 

c 

G 

^ 

^?^ 

E 


GF  EUCLID. 


225 


'is  equal  and  similar  to  the  parallelogram  GF,  and  the  parallelo-BookXI. 
gram  AE  to  BF.     Therefore,  if  a  solid,  8cc.     Q.  E.  D.  *— v— ' 


PROP.  XXV.    THEOR. 

IF  a  solid  parallelepiped  be  cut  by  a  plane  parallel  See  n. 
to  two  of  its  opposite  planes,  it  divides  the  whole  into 
two  solids,  the  base  of  one  of  which  shall  be  to  the         ^ 
base  of  the  other  as  the  one  solid  is  to  the  other. 

Let  the  solid  parallelepiped  ABCD  be  cut  by  the  plane  EV, 
which  is  parallel  to  the  opposite  planes  AR,  HD,  and  divides  the 
whole  into  the  two  solids  ABFV,  EGCD ;  as  the  base  AEFY  of 
the  first  is  to  the  base  EHCF  of  the  other,  so  is  the  solid  ABFV 
to  the  solid  EGCD. 

Produce  AH  both  ways,  and  take  any  number  of  straight  lines 
HM,  MN,  each  equal  to  EH,  and  any  number  AK,  KL,  each 
equal  to  EA,  and  complete  the  parallelograms  LO,  KY,  HQ,  MS, 
and  the  solids  LP,  KR,  HU,  MT;  then,  because  the  straight 
lines  LK,  KA,  AE  are  all  equal,  the  parallelograms  LO,  KY,  AF 


are  equal  a :    and  likewise  the  parallelog»^inis  KX,   KB,  AG  » ;  a  36.  1. 
as  also  ^  the  parallelograms  LZ,  KP,  A^i  because  they  are  op-  b  24. 11. 
posite  planes :  for  the  same  reason,  t.ie  parallelograms  EC,  HQ, 
MS  are  equal  = ;   and  the  parallelr&i'ams  HG,  HI,  IN,  as  also^ 
HD,  MU,  NT :  therefore  three  planes  of  the  solid  LP  are  equal 
and  similar  to  three  planes  of  «:he  solid  KR,  as  also  to  three  planes 
of  the  solid  A  V :    but  the  -'hree  planes  opposite  to  these  three 
are  equal  and  similar  *>  tJ  them  in  the  several  solids,  and  none 
of  their  solid  angles  a-'-e  contained  by  more  than  three  plane  an- 
gles :   therefore  the  three  solids  LP,  KR,   A.V  are  equal  c  to  one  c  C.  11, 
another:   iov  the  same  reason,  the  three  solids  ED,  HU,  MT 
are  equal  to  one  another :   therefore,  what  multiple  soever  the 

2  F 


226 


THE  ELEMENTS 


Book  XL  base  LF  is  of  the  base  AF,  the  same,  multiple  is  the  solid  LV  of 
*--v— ^  the  solid  AV:  for  the  same  reason,  whatever  multiple  the  base 
NF  is  of  the  base  HF,  the  same  multiple  is  the  solid  NV  of  the 
solid  ED :  and  if  the  base  LF  be  equal  to  the  base  NF,  the  solid 
cC.  11  LV  is  equal  =  to  the  solid  NV;  and  if  the  base  LF  be  greater 
than  the  base  NF,  the  solid  LV  is  greater  than  the  solid  NV ; 
and  if  less,  less  :  since  then  there  are  four  magnitudes,  viz.  the 
tv(^o  bases  AF,  FH,  and  the  two  solids  AV,  ED,  and  of  the 

X  B  G        I 


d  5.  def. 
5. 


R 


O  Y  F        C      Q        S 

base  AF  and  solid  AV,  the  base  LF  and  solid  LV  are  any  equi- 
multiples whatever;  and  of  the  base  FH  and  solid  ED,  the  base 
FN  and  solid  N  \^  are  any  equimultiples  whatever ;  and  it  has 
been  proved,  that  if  the  base  LF  is  greater  than  the  base  FN, 
the  solid  LV  is  greater  than  the  solid  NV ;  and  if  equal,  equal ; 
and  if  less,  less.  Therefore  ^  as  the  base  AF  is  to  the  base  FH, 
so  is  the  solid  AV  to  the  solid  ED.  Wherefore,  if  a  solid,  &:c. 
Q.  E.  D. 


PROP.  XXVL     PROB. 


SeeN. 


AT  a  given  point  in  a  given  straight  line,  to  make 
a  solid  angle  equal  to  a  given  solid  angle  contained 
by  three  plane  angles. 


Let  AB  be  a  given  straight  line,  A  a  given  point  in  it,  and  D  a 
given  solid  angle  contained  b;  the  three  plane  angles  EDC,  EDF, 
FDC :  it  is  required  to  make  at  the  point  A,  in  "the  straight  Ime 
AB,  a  solid  angle  equal  to  the  soh-\  anglo  D. 

In  the  straight  line  DF  take  any  point  F,  from  which  draw 

a  11.  11.  a  FG  perpendicular  to  the  plane  ED^,,  meeting  that  plane  in 
G ;    join   DG,   and   at  the   point  A,    in  the  straight  line   AB, 

b  23. 1.     make  ^  the  angle   BAL  equal  to  the   angle  EDC,  and  in   the 
plane  BAL  make  the  angle  BAK  equal  to  the  angle  EDG; 

c  12. 11.  then  make  AK  equal  to  DG,  and  from  the  point  K  erect  ^  KH 


OF  EUCLID. 


227 


at  right  angles  to  the  plane  BAL  ;  and  make  KH  equal  to  Book  XI. 
GF,  and  join  AH :  then  the  solid  angle  at  A,  which  is  contain-  v— y^«ii 
ed  by  the  three  plane  angles  BAL,  BAH,  HAL,  is  equal  to  the 
solid  angle  at  D  contained  by  the  three  plane  angles  EDC,  EDF, 
FDC. 

Take  the  equal  straight  lines  AB,  DE,  and  join  HB,  KB, 
FE,  GE :  and  because  FG  is  perpendicular  to  the  plane  EDC, 
it  makes  right  angles  ^  with  every  straight  line  meeting  it  in  d  3.  def. 
that  plane  :  therefore  each  of  the  angles  FGD,  TGE  is  a  right   ^^^ 
angle :  for  the  same  reason,  HKA,  HKB  are  rTght  angles  :  and 
because    KA,  AB   are   equal    to  GD,   DE,  each   to  each,    and 
contain   equal  angles,  therefore  the  base  PK.  is  equal  «  to  the  e  4. 1. 
base  EG  :  and  KH  is  equal  to  GF,  and  HKB,  FGE  are  right 
angles,  therefore   HB   is   equal  e  to  FF:    again,  because  AK, 
KH   are  •  equal   to   DG,    GF,   and   contain   right   angles,   the 
base  AH  is  equal  to  the  base  DF  :  and  AB  is  equal  to  DE ; 
therefore  HA,  AB  are  equal  to  FD?  DE,  and  the  base  HB  is  e- 
qual  to  tlxe  base  FE,  ~ 

therefore    the    angle 

BAH    is    equal  f   to  //1\  //\\  £8.1, 

the  angle  EDF:  for 
th^  same  reason,  the 
angle  HAL  is  equal 
to  tht  angle  FDC. 
Because  if  AL  and 
DC  be  made  equal, 
and  KL,  HL,  GC, 
FC  be  joined,  since 
whole  EDC,  and  the  parts  of 
the  construction,  equal ;  therefore 
is   equal    to   the    remaining   angle 


B 


the   whole 
parts 


angle 
them 


the 
by 


BAL  is  equal  to 
BAK,  EDG  are, 
the  remaining  angle  KAL 
GDC :  and  because  KA, 
AL  are  equal  to  GD,  DC,  and  contain  equal  angles,  the  base 
KL  is  equal  e  to  the  base  GC  :  and  KH  is  equal  to  GF,  so  that 
LK,  KH  are  equal  to  CG,  GF,  and  they  contain  right  angles ; 
therefore  the  base  HL  is  equal  to  the  base  FC :  again,  because 
HA,  AL  are  equal  to  FD,  DC,  and  the  base  HL  to  the  base 
FC,  the  angle  HAL  is  equal  f  to  the  angle  FDC :  therefore, 
because  the  three  plane  angles  BAL,  BAH,  HAL,  which  contain 
the  solid  angle  at  A,  are  equal  to  the  three  plane  angles  EDC, 
EDF,  FDC,  which  contain  the  solid  angle  at  D,  each  to  each, 
and  are  situated  in  the  same  order,  the  solid  angle  at  A  is  e- 
qual  e  to  the  solid  angle  at  D.  Therefore,  at  a  given  point  in  g  B.  Jl 
a  given  straight  line,  a  solid  angle  has  been  made  equal  to  a  gi- 
ven solid  angle  contained  by  three  plane  angles.  Which  Avas  tp 
be  done. 


THE  ELEMENTS 


PROP.  XXVII.  PROB. 


Nl 


D 


TO  describe  froril  a  given  straight  line  a  solid  pa- 
rallelepiped similar  and  similarly  situated  to  one 
given. 

Let  AB  be  the  g.ven  straight  line,  and  CD  the  given  solid  pa- 
rallelepiped;   it  is  rtquired  from  AB  to  describe  a  solid  paral- 
lelepiped similar  and  sonilarly  situated  to  CD. 
a  26. 11.       At  the  point  A  of  th«t  given  sti-aight  line  AB  make»  a  solid 
angle  equal  to  the  solid  a^gle  at  C,  and  let  BAK,  KAH,  HAB 
be  the  three  plane  angles  '^hich  contain  it,  so  that  BAK  be  e- 
qual  to  the    angle    ECG,    aiid    KAH    to    GCF,    and   HAB    to 
FCE :  and  as  EC  to  CG,  so  mcAce  ^  BA  to  AK ;  and  as  GC  to 
b  12.  6.    CF,   so  make  ^   K  A  to  AH ;    wherefore,  ex  (cquali  c,  as  EC  to 
c  22.  5.    CF,  so  is  BA  to  AH:   complete  the  parallelogram  BH,  and 
the  solid  AL:  and  L 

because    as  EC  to 
CG,  so  BA  to  AK, 
the  sides  about  the 
equal  angles  ECG, 
BAK   are    propor- 
tionals :     therefore 
the    parallelogram 
BK    is   similar    to 
EG.    For  the  same 
reason,  the  parallelogram  KH  is  similar  to  GF,  and  HB  to  FE. 
Wherefore,  three  parallelograms  of  the  solid  AL  are  similar  to 
three  of  (he  solid  CD  ;  and  the  three  opposite  ones  in  eath  solid 
d  24. 11.  are  equaH  and  similar  to  these,  each  to  each.      Also,  because 
the  plane  angles  which  contain  the  solid  angles  of  the  figures 
are  equal,  each  to  each,  and  situated  in  the  same  order,  the  solid 
e  B.  11.    angles  are  equal  *,  each  to  each.      Therefore  the  solid  AL  is 
f  11.  def.  similar  f  to  the  solid  CD.      Wherefore   from   a   given   straight 
11-        line  AB  a  solid  parallelepiped  AL  has  been  described,  similar 
and  similarly  situated  to  the  given  one  CD.     Which  was  to  be 
done. 


K«^ 


\ 

H 

\ 

\ 
\ 

\ 

^ — K 


A 


B 


OF  EUCLID. 


PROP.  XXVIII.    THEOR. 


IF  a  solid  parallelepiped  be  cut  by  a  plane  passing  See  n> 
through  the  diagonals  of  two  of  the  opposite  planes ; 
it  shall  be  cut  in  two  equal  parts. 

Let  AB  be  a  solid  parallelepiped,  and  DE,  CF  the  diagonals 
of  the  opposite  parallelograms  AH,  GB,  viz.  those  which  are 
drawn  betwixt  the  equal  angles  in  each  :  and  because  CD,  FE 
are  each  of  them  parallel  to  GA,  and  not  in  the  same  plane  with 
it,  CD,  FE  are  parallel^;  wherefore  the  diagonals  CF,  DE  are  a  9. 11. 
in  the  plane  in  which  the  parallels  are, 

and  are  themselves  parallels^:  and  the  \^ ~         b  16.  11. 

plane  CDEF  shall  cut  the  solid  AB  in- 
to tv/o  equal  parts. 

Because  the  triangle  CGF  is  equal  <^        1  ""  c  34. 1- 
to  the  triangle  CBF,   and  the  triangle 
DAE    to   DHE:   and   that   the    paral- 
lelogram CA  is  equaH  and   similar  to        L^Jtr— ■ !■ — JH    d  24. 11. 

the  opposite  one  BE ;  and  the  paral- 
lelogram GE  to  CH :  therefore  the 
prism  contained  by  the    two  triangles 

CGF,  DAE,  and  the  three  parallelograms  CA,  GE,  EC,  is  equal  e  e  C  11. 
to  the  prism  contained  by  the  two  triangles  CBF,  DHE,  and  the 
three  parallelograms  BE,  CH,  EC,  because  they  are  contained 
by  the  same  number  of  equal  and  similar  planes,  alike  situated, 
and  none  of  their  solid  angles  are  contained  by  more  than  three 
plane  angles.  Therefore  the  solid  AB  is  cut  into  two  equal  parti 
by  the  plane  CDEF.     Q.  E.  D. 

'  N.  B.  The  insisting  straight  lines  of  a  parallelepiped,  men- 
'  tioned  in  the  next  and  some  folio wingpropositions,  are  the  sides 
'  of  the  parallelograms  betwixt  the  base  and  the  opposite  plane 
'  parallel  to  it.' 


c 

B 

o^ 

D 

F 

A 

/ 

PROP.  XXIX.     THEOR. 


SOLID  parallelepipeds  upon  the  same  base,  and  of  See  x 
the  same  altitude,  the  insisting  straight  lines  of  which 
are  terminated  in  the  same  straight  lines  in  the  plane 
opposite  to  the  base,  are  equal  to  one  another. 


230 


THE  ELEMENTS 


See  the 

figures 

below. 


Book XI.  Let  the  solid  parallelepipeds  AH,  AK  be  upon  the  same  base 
AB,  and  of  the  same  altitude,  and  let  their  insisting  straight  lines 
AP\  AG,  LM,  LN  be  terminated  in  the  same  straight  line  FN, 
and  CD,  CE,  BH,  BK  be  terminated  in  the  same  straight  line 
DK ;  the  solid  AH  is  equal  to  the  solid  AK. 

First,  let  the  parallelograms  DG,  UN,  which  are  opposite  to 
the  base  y\.B,  have  a  common  side  HG  :  then,  because  the  solid 
AH  is  cut  by  the  plane  AGHC  passing  tn rough  the  diagonals 
AG,  CH  of  the  opposite  planes  ALGF,  CB'ilD,  AH  is  cut  into 

a  28. 11.  two  equal  parts  »  by  the  plane  AGHC  :  therefore  the  solid  AH  is 
double  of  the   prism   which  is  contained  betwixt  the  triangles 
ALG,  CBH:  for  the  same  rea- 
son, because  the   solid   AK  is 
cutby  the  plane  LGHB  through  ^ 

the  diagonals  LG,  BH  of  the  op-  ^  ^    ^  I  ^  ^^ ^  N 

posite  planes  ALNG,  CBKH, 
the  solid  AK  is  double  of  the 
same  prism  which  is  contain- 
ed betwixt  the  triangles  ALCi, 
CBH.  Therefore  the  solid  AH  ^  L 

is  equal  to  the  solid  AK. 

But,  let  the  parallelograms  DM,  EN  opposite  to  the  base, 
have  no  common  side  :  then,  because  CH,  CK  are  parallelo- 
grams, CB  is  equal  i>  to  each  of  the  opposite  sides  DH,  EK ; 
wherefore  DH  is  equal  to  EK :  add,  or  take  away  the  common 
part  HE;  then  DE  is  equal  to  HK :  wherefore  also  the  tri- 
angle CDE  is  equal  c  to  the  triangle  BHK :  and  the  parallelo- 
gram DG  is  equal  ti  to  the  parallelogram  HN :  for  the  same 
reason,  the  triangle  AFG  is  equal  to  the  triangle  LMX,  and 
the  parallelogram  CF  is  equal  e   to  the  parallelogram  BM,  and 


D 

H 

K 

\F 

^fV!/" 

r 

/ 

7' 

/ 

^ 

^  / 

\ 

/ 

1 

/ 

b  34.  1. 

c  38.  1. 
d  36.  1. 

e  24-  11 
D 


CG  to  BN  ;  for  they  are  opposite,  Therefore  the  prism  which 
is  contained  by  the  two  triangles  AFG,  CDE,  and  the  three 
parallelograms  AD,  DG,  GC,  is  equal f  to  the  prism  contain- 
t  C  11.  <-'d  by  tlie  two  triangles  LMN%  BHK,  and  the  three  parallelo- 
grams  BM,   IMK,  KL.     If  therefore  the  prism  Lr.INBHK  bp 


OF  EUCLID.  231 

taken  from  the  solid  of  which  the  base  is  the  parallelogram  AB,  Book XI. 
and  in  which  FDKN  is  the  one  opposite  to  it ;    and  if  from  this  *«— v-^* 
same  solid  there  be  taken  the  prism  AFGCDE,  the  remaining 
solid,  viz.  the  parallelepiped  AH,  is  equal  to  the  remaining  pa- 
rallelepiped AK.    Therefore,  solid  parallelepipeds,  8cc.    Q.  E.  D. 

PROF.  XXX.     THEOR. 

SOLID  parallelepipeds  upon  the  same  base,  and  See  n 
of  the  same  altitude,   the  insisting  straight  lines  of 
which  are  not  terminated  in  the  same  straight  lines 
in  the  plane  opposite  to  the  base,  are  equal  to  one 
another. 

Let  the  parallelepipeds  CM,  CN  be  upon  the  same  base  AB, 
and  of  the  same  altitude,  but  their  insisting  straight  lines  AF, 
AG,  LM,  LN,  CD,  CE,  BH,  BK  not  terminated  in  the  same 
straight  lines :  the  solids  CM,  CN  are  equal  to  one  another. 

Produce  FD,  MH,  and  NG,  KE,  and  let  them  meet  one  ano- 
ther in  the  points  O,  P,  Q,  R ;  and  join  AO,  LP,  BQ,  CR :  and 
because  the  plane  LBHM  is  parallel  to  the  opposite  plane  ACDF, 

N  K 


M  H 


and  that  the  plane  LBHM  is  that  in  which  are  the  parallels  LB, 
MHPQ,  in  which  also  is  the  figure  BLPQ  ;  and  the  plane  ACDF 
is  that  in  which  are  the  parallels  AC,  FDOR,  in  which  also  is 
the  figure  C  AOR ;  therefore  the  figures  BLPQ,  C  AOR  are  in 
parallel  planes :  in  like  manner,  because  the  plane  ALNG  is  pa- 
rallel CO  the  opposite  plane  CBKE,  and  that  the  plane  ALNG  is 
that  in  which  are  the  parallels  AL,  OPGN,  in  which  also  is  the 


9,32 


THE  ELEMENTS 


Book XI.  figure  ALPO  ;  and  the  plane  CBKE  is  that  in  which  are  the  pa- 
'^— V— '  rallels  CB,  RQEK,  in  which  also  is  the  figure  CBQR ;  therefore 
the  figures  ALPO,  CBQR  are  in  parallel  planes:  and  the  planes 
ACBL,  ORQP  are  parallel ;  therefore  the  solid  CP  is  a  paralle- 
lepiped :  but  the  solid  CM,  of  which  the  base  is  ACBL,  to  which 
a  29. 11.  FDHM  is  the  opposite  parallelogram,  is  equal  ^  to  the  solid  CP, 
of  which  the  base  is  the  parallelogram  ACBL,  to  which  ORQP 


A  C 

is  the  one  opposite,  because  they  are  apon  the  same  base,  and 
their  insisting  straight  lines  AF,  AO,  CD,  CR ;  LM,  LP,  BH, 
BQ  are  in  the  same  straight  lines  FR,  MQ :  and  the  solid  CP  is 
equal  »  to  the  solid  CN  ;  for  they  are  upon  the  same  base  ACBL, 
and  their  insisting  straight  lines  AO,  AG,  LP,  LN ;  CR,  CE, 
BQ,  BK  are  in  the  same  straight  lines  OL,  RK;  therefore  the 
solid  CM  is  equal  to  the  solid  CN.  Wherefore,  solid  parallelepi- 
peds, &c.     Q.  E.  D. 


PROP.  XXXL     THEOR. 


SeeN. 


SOLID  parallelepipeds  which  are  upon  equal  bases, 
and  of  the  same  altitude,  are  equal  to  one  another. 


Let  the  solid  parallelepipeds  AE,  CF  be  upon  equal  bases  AB, 
CD,  and  be  of  the  same  altitude  ;  the  solid  AE  is  equal  to  the  so- 
lid CF. 

First,  Let  the  insisting  straight  lines  be  at  right  angles  to  the 
bases  AB,  CD,  and  let  the  bases  be  placed  in  the  same  plane, 


OF  EUCLID. 


333 


and  so  as  that  the  sides  CL,  LB  be  in  a  straight  line ;  therefore  Book  XL 
the  straight  line  LM,  which  is  at  right  angles  to  the  plane  in  ' — r— ' 
which  the  bases  are,  in  the  point  L,  is  common  »  to  the  two  so-  a  13. 11. 
lids  AE,  CF :  let  the  other  insisting  lines  of  the  solids  be  AG, 
HK,  BE;   DF,  OP,  CN :  and  first  let  the  angle  ALB  be  equal 
to  the  angle  CLD  ;  then  AL,  LD  are  in  a  straight  linet.     Pro- b  14.1. 
duce  OD,  HB,  and  let  them  meet  in  Q,  and  complete  the  solid 
parallelepiped  LR,  the  base  of  which  is  the  parallelogram  LQ, 
and  of  which  LM  is  one  of  its  insisting  straight  lines  :  therefore, 
because  the  parallelogram  AB  is  equal  to  CD,  as  the  base  AB 
is  to  the  base  LQ,  so  is  <=  the  base  CD  to  the  same  LQ :  and  be-  c  7.  5. 
cause  the  solid  parallelepiped  AR  is  cut  by  the  plane  LMEB, 
which  is  parallel  to  the  opposite  planes  AK,  DR ;  as  the  base 


AB  is  to  the  base  LQ,   so  is 


d  the  solid  AE  to  the  solid  LR  :  d  25. 11. 


for   the   same    reason,  because   the   solid   parallelepiped  CR  is 
cut  by  the  plane  LMFD,  which  is  parallel  to  the  opposite  planes 


R 


O 


"" 

NT 

\iVl 

\ 

E 

0 

^ 

■v^V 

7 

^ 

p-^ 

G 

K 

B 

K 

\ 

\ 

c 

L 

^ 

h-, 

^ 

CP,  BR ;  as  the 
base  CD  to  the  base 
LQ,  so  is  the   solid 

CF  to  the  solid  LR :  xr-^j         |  "^^^^T^  X 

but  as  the  base  AB  '^ 

to  the  base  LQ,  so 
the  base  CD  to  the 
base   LQ,  as  before 
was  proved:  there- 
fore as  the  solid  AE  a      c  H     T 
to  the  solid  LR,  so                                      ^     ^  "      ^ 
is  the  solid  CF  to  the  solid  LR ;  and  therefore  the  solid  AE  is 
equal  e  to  the  solid  CF.  *  9-  ^• 

But  let  the  solid  parallelepipeds  SE,  QF  be  upon  equal  bases 
SB,  CD,  and  be  of  the  same  altitude,  and  let  their  insisting 
straight  lines  be  at  right  angles  to  the  bases  ;  and  place  the  bases 
SB,  CD  in  the  same  plane,  so  that  CL,  LB  be  in  a  straight  lincj 
and  let  the  angles  SLB,  CLD  be  unequal ;  the  solid  SE  is  also 
in  this  case  equal  to  the  solid  CF :  produce  DL,  TS  until  they 
meet  in  A,  and  from  B  draw  BH  parallel  to  DA ;  and  let  HB, 
OD  produced  meet  in  Q,  and  complete  the  solids  AE,  LR : 
therefore  the  solid  AE,  of  which  the  base  is  the  parallelogram 
LE,  and  AK  the  one  opposite  to  it,  is  equal*"  to  the  solid  SE,  off  29. 11. 
which  the  base  is  LE,  and  to  which  SX  is  opposite  ;  for  they  are 
upon  the  same  base  LE,  and  of  the  same  altitude,  and  their  in- 
sisting straight  lines,  viz.  LA,  LS,  BH,  BT ;  MG,  MV,  EK, 
EX  are  in  the  same  straight  lines  AT,  GX :  and  because  the 

2  G 


234 


THE  ELEMENTS 


Book  XI.  parallelogram  AB  is  equal  k  to  SB,  for  they  are  upon  the  same 

<-.  y— ^  base  LB,  and  between  the  same  parallels  LB,  AT  ;  and  that  the 

g  35. 1.  base  SB  is  equal  to 
the  base  CD  ;  there- 
fore the  base  AB  is 
equal  to  the  base  CD, 
and  the  angle  ALB 
is  equal  to  the  angle  O 
CLD:  therefore,  by 
the  first  case,  the  so- 
lid AE  is  equal  to  the 
solid  CF  ;  but  the  so- 
lid AE  is  equal  to  the 
solid  SE,  as  was  demonstrated ;  therefore  the  solid  SE  is  equsl 
to  the  solid  CF. 

But  if  the  insisting  straight  lines  AG,  HK,  BE,  LM ;  CN, 
RS,  DF,  op  be  not  at  right  angles  to  the  bases  AB,  CD ;  in 
this  case  likewise,  the  solid  AE  is  equal  to  the  solid  CF :  from 
the  points  G,  K,  E,  M ;  N,  S,  F,  P  draw  the  straight  lines  GQ, 

hll.  11.  KT,  EV,  MX;  NY,  SZ,  FI,  PU,  perpendicular ^  to  the  plane 
in  which  are  the  bases  AB,  CD ;  and  let  them  meet  it  in  the 
points  Q,  T,  V,  X ;  Y,  Z,  I,  U,  and  join  QT,  TV,  VX,  XQ ; 
YZ,  ZI,  lU,  UY :  then,  because  GQ,  KT  are  at  right  angles  to 


Q        T  C  R  Y  Z 

i  6. 11.  the  same  plane,  they  are  parallel '  to  one  another :  and  MG, 
EK  are  parallels ;  therefore  the  plane  MQ,  ET,  of  which  one 
passes  through  MG,  GQ,  and  the  other  through  EK,  KT,  which 
are  parallel  to  MG,  GQ,  and  not  in  the  same  plane  with  them> 

kl5. 11.  are  parallel  J'  to  one  another:  for  the  same  reason,  the  planes 
MV,  GT  are  parallel  to  one  another :  therefore  the  solid  QE  is 
a  parallelepiped  :  in  like  manner,  it  may  be  proved,  that  the  SO'- 
lid  YF  is  a  parallelepiped :  but,  from  what  has  been  domonstrat- 
ed,  the  solid  EQ  is  equal  to  the  solid  FY,  because  they  are  upon 
equal  bases  MK,  PS,  and  of  the  same  altitude,  and  have  their 
insisting  straight  lines  at  right  angles  lo  the  bases:  and  the  so- 


OF  EUCLID, 


23-5 


lid  EQ  is  equal  1  to  the  solid  AE ;  and  the  solid  FY,  to  the  solid  Book  XI. 
CF :  because  they  are  upon  the  same  bases  and  of  the  same  alti-  ^— v^^ 
tude :  therefore  the  solid  AE  is  equal  to  the  solid  CF.     Where- 1  29.  or 
fere,  solid*t)arallelepipeds,  &c.    Q.  E.  D.  30. 11 


PROP.  XXXII.     THEOR. 

SOLID  parallelepipeds  which  have  the  same  alti-  See  n. 
tude,  are  to  one  another  as  their  bases. 

Let  AB,  CD  be  solid  parallelepipeds  of  the  same  altitude :  they 
are  to  one  another  as  their  bases ;  that  is,  as  the  base  AE  to  the 
base  CF,  so  is  the  solid  AB  to  the  solid  CD. 

To  the  straight  line  FG  apply  the  parallelogram  FH  equal  =i  a  Cor.  45. 
to  AE,  so  that  the  angle  FGH  be  equal  to  the  angle  LCG,  and   ^• 
complete  the  solid  parallelepiped  GK  upon  the  base  FH,  one  of 
whose  insisting  lines  is  FD,  whereby  the  solids  CD,  GK  must  be 
of  the  same  altitude:  therefore  the  solid  AB  is  equal''  to  thebis. ll- 
solid  GK,  because 

they   are  upon  e-  B  D  K 

qual  bases  AE, 
FH,  and  are  of 
the  same  altitude : 
and  because  the 
solid  parallelepi- 
ped CK  is  cut  by 
the  plane  DG, 
which   is  parallel 

to  its  opposite  planes,  the  base  HF  is^  to  the  base  FC,  as  the  so- c  25. 11 
lid  HD  to  the  solid  DC :  but  the  base  HF  is  equal  to  the  base  AE, 
and  the  solid  GK  to  the  solid  AB :  therefore,  as  the  base  AE  to 
the  base  CF,  so  is  the  solid  AB  to  the  solid  CD.  Wherefore, 
solid  parallelepipeds,  &c.     Q.  E.  D. 

Cor.  From  this  it  is  manifest  that  prisms  upon  triangular 
bases,  of  the  same  altitude,  are  to  one  another  as  their  bases. 

Let  the  prisms,  the  bases  of  which  are  the  triangles  AEMj 
CFG,  and  NBO,  PDQ  the  triangles  opposite  to  them,  have  the 
same  altitude ;  and  complete  the  parallelograms  AE,  CF,  and 
the  solid  parallelepipeds  AB,  CD,  in  the  first  of  which  let  MO, 
and  in  the  other  let  GQ  be  one  of  the  insisting  lines.  And  be- 
cause the  solid  parallelepipeds  AB,  CD  have  the  same  altitude, 
they  are  to  one  another  as  the  base  AE  is  to  the  base  CF  j  where- 


236 


THE  ELEMENTS 


Book  XI.  fore  the  prisms,  whicli  are  their  halves^,  are  to  one  another  as 
^-"■v— ^  the  base  AE  to  the  base  CF ;  that  is,  as  the  triangle  AEM  to  the 
d  28. 11.  triangle  CFG.  ^ 


PROP.  XXXIII.     THEOR. 

SIMILAR  solid  parallelepipeds  are  one  to  another 
in  the  triplicate  ratio  of  their  homologous  sides. 

Let  AB,  CD  be  similar  solid  parallelepipeds,  and  the  side  AE 
homologous  to  the  side  CF :  the  solid  AB  has  to  the  solid  CD 
the  triplicate  ratio  of  that  which  AE  has  to  CF. 

Produce  AE,  GE,  HE,  and  in  these  produced  take  EK  equal 
to  CF,  EL  equal  to  FN,  and  EM  equal  to  FR  ;  and  complete  the 
parallelogram  KL,  and  the  solid  KG  :  because  KE,  EL  are  equal 
to  CF,  FN,  and  the  angle  KEL  equal  to  the  angle  CFN,  because 
it  is  equal  to  the  ang;le  AEG,  which  is  equal  to  CFN,  by  reason 
that  the  solids  AB,  CD  are  similar;  therefore  the  parallelogram 
KL  is  similar  and  equal  to  tlie  parallelogram  CN :  for  the  same 
reason,  the  parallelogram  MK  is  similar  and  equal  to  CR,  and 
also    OE   to    FD.  -^ 

O 


Therefore  three 
parallelograms  of 
the  solid  KO  are 
equal  and  similar 
to  three  parallelo- 
grams of  the  so- 
lid CD  :  and  the 
three  opposite  ones 
in    each    solid   are 

a24. 11.  equal*  and  simi- 
lar to  these:  there- 
fore  the   solid   KO 

bC.  11.  is  equally  and  si- 
milar to  the  solid  CD 


\ 

\ 

N 

\ 

\ 

R 


h1\p 


A 


E 


M 


\^ 


^ 


O 


cl.6. 


complete  the  parallelogram  GK,  and  com- 
plete the  solids  EX,  LP  upon  the  bases  GK,  KL,  so  that  EH  be 
an  insisting  straight  line  in  each  of  them,  whereby  they  must  be 
of  the  same  altitude  with  the  solid  AB:  and  because  the  solids 
AB,  CD  are  similar,  and,  by  permutation,  as  AE  is  to  CF,  so  is 
EG  to  FN,  and  so  is  EH  to  FR :  and  FC  is  equal  to  EK,  and 
FN  to  EL,  and  FR  to  EM ;  therefore  as  AE  to  EK,  so  is  EG  to 
EL,  and  so  is  HE  to  EM :  but,  as  AE  to  EK,  so^  is  the  paral- 
lelogram AG  to  the  parallelogram  GK  i  and  as  GE  to  EL,  so  is<= 


OF  EUCLID.  237 

GK  to  KL ;  and  as  HE  to  EM,  so  «  is  PE  to  KM  :  therefore  as  Book  XI. 
the  parallelogram  AG  to  the  parallelogram  GK,  so  is  GK  to  KL,  * — r— ' 
and  PE  t^vM  :  but  as  AG  to  GK,  so^  is  the  solid  AB  to  the  c  1.  6. 
solid  EX^nd  as  GK  to  KL,  so*  is  the  solid  EX  to  the  solid  d  25. 11. 
PL ;  and  as  PE  to  KM,  so  ^  is  the  solid  PL  to  the  solid  KO  ;  and 
therefore  as  the  solid  AB  to  the  solid  EX,  so  is  EX  to  PL,  and 
PL  to  KO :  but  if  four  magnitudes  be  continual  proportionals, 
the  first  is  said  to  have  to  the  fourth  the  triplicate  ratio  of  that 
which  it  has  to  the  second:  therefore  the  solid  AB  has  to  the 
solid  KO  the  triplicate  ratio  of  that  which  AB  has  to  EX:    but 
as  AB  is  to  EX,  so  is  the  parallelogram  AG  to  the  parallelogram 
GK,  and  the  straight  line  AE  to  the  straight  line  EK.     Where- 
fore the  solid  AB  has  to  the  solid  KO,  the  triplicate  ratio  of  that 
which  AE  has  to  EK.     And  the  solid  KO  is  equal  to  the  solid 
CD,  and  the  straight  line  EK  is  equal  to  the  straight  line  CF. 
Therefore  the  solid  AB  has  to  the  solid  CD,  the  triplicate  ratio 
of  that  which  the  side  AE  has  to  the  homologous  side  CF,  &c. 
Q.  E.  D. 

CoR.  From  this  it  is  manifest,  that,  if  four  straight  lines  be 
continual  proportionals,  as  the  first  is  to  the  fourth,  so  is  the  so- 
lid parallelepiped  described  from  the  first  to  the  similar  solid 
similarly  described  from  the  second;  because  the  first  straight 
line  has  to  the  fourth  the  triplicate  ratio  of  that  which  it  has  to 
the  second. 


PROP.  D.    THEOR. 

SOLID  parallelepipeds  contained  by  parallelograms  See  n,. 
equiangular  to  one  another,  each  to  each,  that  is,  of 
which  the  solid  angles  are  equal,  each  to  each,  have  . 
to  one  another  the  ratio  which  is  the  same  with  the 
ratio  compounded  of  the  ratios  of  their  sides. 


Let  AB,  CD  be  solid  parallelepipeds,  of  which  AB  is  contain- 
ed by  the  parallelograms  AE,  AF,  AG,  equiangular,  each  to 
each,  to  the  parallelograms  CH,  CK,  CL,  which  contain  the  so- 
lid CD.  The  ratio  which  the  solid  AB  has  to  the  solid  CD  is 
the  same  with  that  which  is  compounded  of  the  ratios  of  the 
aides  AM  to  DL,  AN  to  DK,  and  AO  to  DH. 


238 


THE  ELEMENTS 


Book  XI.     Produce  MA,  NA,  OA  to  P,  Q,  R,  so  that  AP  be  equal  to 

^■^v^-^  DL,  AQ  to  DK,  and  AR  to  DH ;  and  complete  the  solid  paral- 
lelepiped AX  contained  by  the  parallelograms  AS,  AT,  AV  si- 
milar and  equal  to  CH,  CK,  CL,  each  to  each.     TBw-efore  the 

a  C.  11.  solid  AX  is  equal  *  to  the  solid  CD.  Complete  likewise  the  solid 
AY,  the  base  of  which  is  AS,  and  of  which  AO  is  one  of  its  in- 
sisting straight  lines.  Take  any  straight  line  a,  and  as  MA  t» 
AP,  so  make  a  to  b ;  and  as  NA  to  AQ,  so  make  b  to  c ;  and  as 
AO  to  AR,  so  c  to  d :  then,  because  the  parallelogram  AE  is 
equiangular  to  AS,  AE  is  to  AS,  as  the  straight  line  a  to  c,  as 
is  demonstrated  in  the  23d  prop,  book  6,  and  the  solids  AB,  AY, 
being  betwixt  the  parallel  planes  BOY,  EAS,  are  of  the  same 

b  32  11.  altitude.  Therefore  the  solid  AB  is  to  the  solid  AY,  as  ^  the 
base  AE  to  the  base  AS ;  that  is,  as  the  straight  line  a  is  to  c. 

c  25. 11.  And  the  solid  AY  is  to  the  solid  AX,  as  <=  the  base  OQ  is  to  the 


D  L 


B 


li 


\ 

^ 

K 

\ 

\ 

E 

c 

C 

P 

\" 

N 

\ 

\ 

P 

\ 

. 

\ 

M 

A 

Q 

R 

T 

\ 

\ 

V  X 


base  QR ;  that  is,  as  the  straight  line  OA  to  AR ;  that  is,  as  the 
straight  line  c  to  the  straight  line  d.  And  because  the  solid  AB 
is  to  the  solid  AY,  as  a  is  to  c,  and  the  solid  AY  to  the  solid  AX 
as  c  is  to  d ;  ex  ctquali^  the  solid  AB  is  to  the  solid  AX, 
or  CD  which  is  equal  to  it,  as  the  straight  line  a  is  to  d.  But 
4  def.  A.  the  ratio  of  a  to  d  is  said  to  be  compounded  "^  of  the  ratios  of  a  to  b, 
^-  b  to  c,  and  c  to  d,  which  are  the  same  with  the  ratios  of  the  sides 
MA  to  AP,  NA  to  AQ,  and  OA  to  AR,  each  to  each.  And  the 
sides  AP,  AQ,  AR  are  equal  to  the  sides  DL,  DK,  DH,  each  to 
each.  Therefore  the  solid  AB  has  to  the  soHd  CD  the  ratio 
which  is  the  same  with  that  which  is  compounded  of  the  ratios 
of  the  sides  AM  to  DL,  AN  to  DK,  and  AO  to  DH.    Q.  E,  D. 


OF  EUCLID. 


PROP.  XXXIV.    THEOR. 


THE  bases  and  altitudes  of  equal  solid  parallelepi-  See  n. 
peds  are  reciprocally  proportional ;   and,  if  the  bases 
and  altitudes  be  reciprocally  proportional,  the  solid 
parallelepipeds  are  equal. 


Let  AB,  CD  be  equal  solid  parallelepipeds ;  their  bases  are 
reciprocally  proportional  to  their  altitudes;  that  is,  as  the  base 
EH  is  to  the  base  NP,  so  is  the  altitude  of  the  solid  CD  to  the 
altitude  of  the  solid  AB. 

First,  Let  the  insisting  straight  lines  AG,  EF,  LB,  HK ;  CM, 
NX,  OD,  PR  be  at  right  angles  to  the  bases.  As  the  base  EH 
to  the  base  NP,  so  is  CM  to 

K  B 


R 


D 


FT 


H 


M 


,0 


X 


N 


AG.  If  the  base  EH  be 
equal  to  the  base  NP,  then 
because  the  solid  AB  is  like- 
wise equal  to  the  solid  CD, 
CM  shall  be  equal  to  AG. 
Because  if  the  bases  EH, 
NP  be  equal,  but  the  alti- 
tudes AG,  CM  be  not  equal, 
neither  shall  the  solid  AB 
be  equal  to  the  solid  CD.  But  the  solids  are  equal,  by  the  hypo- 
thesis. Therefore  the  altitude  CM  is  not  unequal  to  the  altitude 
AG  ;  that  is,  they  are  equal.  Wherefore,  as  the  base  EH  to  the 
base  NP,  so  is  CM  to  AG. 

Next,  Let  the  bases  EH,  NP  not  be  equal,  but  EH  greater 
than  the  other :  since  then 
the  solid  AB  is  equal  to 
the  solid  CD,  CM  is 
therefore  greater  than 
AG ;  for,  if  it  be  not, 
neither  also  in  this  case 
would  the  solids  AB,  CD 
be  equal,  which,  by  the 
hypothesis,  are  equal. 
Make  then  CT  equal  toH 
AG,  and  complete  the 
solid  parallelepiped  CV, 


K 


B 


E 


240 


THE  ELEMENTS 


Book XI.  of  which  the  base  is  NP,  and  altitude  CT.  Because  the  solid 
AB  is  equal  to  the  solid  CD,  therefore  the  solid  AB  is  to  the  so- 
lid CV,  as  a  the  solid  CD  to  the  solid  CV.  But  as  the  solid  AB 
to  the  solid  CV,  so''  is  the  base  EH  to  the  base  NP";  for  the  so- 
lids AB,  C'V  are  of  the  same  altitude;  and  as  the  solid  CD  to 
C  V,  so  c  is  the  base  MP  to  the  base  PT,  and  so  ^  is  the  straight 
line  MC  to  CT ;  and  CT  is  equal  to  AG.  Therefore,  as  the 
base  EH  to  the  base  NP,  so  is  MC  to  AG.  Wherefore  the  bases 
of  the  solid  parallelepipeds  AB,  CD  are  reciprocally  proportional 
to  their  altitudes. 

Let  now  the  bases  of  the  solid  parallelepipeds  AB,  CD  be  re- 
ciprocally proportional  to  their  altitudes,  viz.  as  the  base  EH 
to  the  base  NP,  so  the  alti- 


c  25. 11. 
dl.  6. 


tude  of  the  solid  CD  to  the  K 
altitude  of  the  solid  AB ; 
the  solid  AB  is  equal  to 
the  solid  CD.  Let  the  in- 
sisting lines  be,  as  before, 
at  right  angles  to  the  bases. 
Then,  if  the  base  EH  be  H 
equal  to  the  base  NP,  since 
EH  is  to  NP,  as  the  alti- 
tude of  the  solid  CD  is  to  the 


B 


R 


\( 

ur 

\ 

L 

\ 

\ 

M 


D 

K 


E 


N 


is  to  tne  altitude  of  the  solid  AB,  there- 
e  A.  5.     fore  the  altitude  of  CD  is  equal  «  to  the  altitude  of  AB.     But  so- 
lid parallelepipeds  upon  equal  bases,  and  of  the  same  altitude, 
f  31. 11.  are  equal  f  to  one  another :    therefore  the  solid  AB  is  equal  to 
the  solid  CD. 

But  let  the  bases  EH,  NP  be  unequal,  and  let  EH  be  the 
greater  of  the  two.     Therefore,  since  as  the  base  EH  to  the  base 
NP,  so  is  CM  the  alti- 
tude of  the  solid  CD  to 
AG  the  altitude  of  AB, 
CM  is   greater  e  than  B 

AG.  Again,  take  CT  K 
equal  to  AG,  and  com- 
plete, as  before,  the  so- 
lid CV.  And  because 
the  base  EH  is  to  the  H 
base  NP,  as  CM  to  AG, 
and  that  AG  is  equal  to  j^ 

C T,  therefore  the  base 
EH  is  to  the  base  NP,  as  MC  to  CT.  But  as -the  base  EH  is  to  NP, 
so  •>  is  the  solid  AB  to  the  solid  CV  ;  for  the  solids  AB,  CV  are  of 
the  same  altitude ;  and  as  MC  to  CT,  so  is  the  base  MP  to  the  base 


G 

I. 

OF  EUCLID. 


241 


PT,  and  the  solid  CD  to  the  solid  ^CV:  and  therefore  as  the  BookXf, 
solid  AB  to  the  solid  C V,  so  is  the  solid  CD  to  the  solid  CV ;  v—y—^ 
that  is,  each  of  the  solids  AB,  CD  has  the  same  ratio  to  the  solid  c  25. 11. 
CV ;  and  therefore  the  solid  AB  is  equal  to  the  solid  CD. 

Second  general  case.  Let  the  insisting  straight  lines  FE, 
BL,  GA,  KH;  XN,  DO,  MC,  RP  not  be  at  right  angles  to 
the  bases  of  the  solids ;  and  from  the  points  F,  B,  K,  G ;  X, 
D,  R,  M  draw  perpendiculars  to  the  planes  in  which  are  the 
bases  EH,  NP  meeting  those  planes  in  the  points  S,  Y,  V,  T  ; 
Q,  I,  U,  Z ;  and  complete  the  solids  FV,  XU,  which  are  pa- 
rallelepipeds, as  was  proved  in  the  last  part  of  prop.  31,  of 
this  book.  In  this  case,  likewise,  if  the  solids  AB,  CD  be 
equal,  their  bases  are  reciprocally  proportional  to  their  altitudes, 
viz.  the  base  EH  to  the  base  NP,  as  the  altitude  of  the  solid 
CD  to  the  altitude  of  the  solid  AB.  Because  the  solid  AB  is 
equal  to  the  solid  CD,  and  that  the  solid  BT  is  equal  s  to  the  g  29.  qr 
solid  BA,  for  they  are  upon  the   same   base  FK,  and  of  the  30. 11. 


same .  altitude ;  and  that  the  solid  DC  is  equal  s  to  the  solid 
DZ,  being  upon  the  same  base  XR,.  and  of  the  same  altitude ; 
therefore  the  solid  BT  is  equal  to  the  solid  DZ  :  but  the  bases 
are  reciprocally  proportional  to  the  altitudes  of  equal  solid  pa- 
rallelepipeds of  which  the  insisting  straight  lines  are  at  right 
angles  to  their  bases,  as  before  was  proved  :  therefore  as  the 
base  FK  to  the  base  XR,  so  is  the  altitude  of  the  solid  DZ  to 
the  altitude  of  the  solid  BT :  and  the  base  FK  is  equal  to  the 
base  EH,  and  the  base  XR  to  the  base  NP :  wherefore  as  the 
base  EH  to  the  base  NP,  so  is  the  altitude  of  the  solid  DZ  to 
the  altitude  of  the  solid  BT :  but  the  altitudes  of  the  solids 
DZ,  DC,  as  also  of  the  sohds  BT,  BA,  are  the  same.  There- 
fore as  the  base  EH  to  the  base  NP,  so  is  the  altitude  of  the 

2H 


242 


THE  ELEMENTS 


Book  XI.  solid  CD  to  the  altitude  of  the  solid  AB ;  that  is,  the  bases  of  the 
^-^■-^  solid  parallelepipeds  AB,  CD  are  reciprocally  proportional  to  their 
altitudes. 

Next,  Let  the  bases  of  the  solids  AB,  CD  be  reciprocally  pro- 
portional to  their  altitudes,  viz.  the  base  EH  to  the  base  NP,  as  the 
altitude  of  the  solid  CD  to  the  altitude  of  the  solid  AB  ;  the  solid 
AB  is  equal  to  the  solid  CD  :  the  same  construction  being  made  ; 
because,  as  the  base  EH  to  the  base  NP,  so  is  the  altitude  of  the 
solid  CD  to  the  altitude  of  the  solid  AB ;  and  that  the  base  EH  is 
equal  to  the  base  FK  ;  and  NP  to  XR ;  therefore  the  base  FK  is 
to  the  base  XR,  as  the  altitude  of  the  solid  CD  to  the  altitude  of 


Pv 


D 


g  29.  o* 
30.  11. 


AB :  but  the  altitudes  of  the  solids  AB,  BT  are  the  same,  as  also 
of  CD  and  DZ ;  therefore  as  the  base  FK  to  the  base  XR,  so  is 
the  altitude  of  the  solid  DZ  to  the  altitude  of  the  solid  BT  t 
wherefore  the  bases  of  the  solids  BT,  DZ  are  reciprocally  propor- 
tional to  their  altitudes ;  and  their  insisting  straight  lines  are  at 
right  angles  to  the  bases ;  wherefore,  as  was  before  proved,  the 
solid  BT  is  equal  to  the  solid  DZ :  but  BT  is  equal  g  to  the  solid 
BA,  and  DZ  to  the  solid  DC,  because  they  are  upon  the  same 
bases,  and  of  the  same  altitude.  1  herefore  the  solid  AB  is  equal 
to  the  solid  CD.     Q.  E.  D. 


OF  EUCLID. 


PROP.  XXXV.    THEOR. 


IF  from  the  vertices  of  two  equal  plane  angles  there  See  N. 
be  drawn  two  straight  lines  elevated  above  the  planes 
in  which  the  angles  are,  and  containing  equal  angles 
with  the  sides  of  those  angles,  each  to  each ;  and  if 
in  the  lines  above  the  planes  there  be  taken  any  points, 
and  from  them  perpendiculars  be  drawn  to  the  planes 
in  which  the  first-named  angles  are;  and  from  the 
points  in  which  they  meet  the  planes,  straight  lines 
be  drawn  to  the  vertices  of  the  angles  first- named ; 
these  straight  lines  shall  contain  equal  angles  with  the 
straight  lines  which  are  above  the  planes  of  the  angles. 

Let  BAC,  EDF  be  two  equal  plane  angles;  and  from  the 
points  A,  D  let  the  straight  lines  AG,  DM  be  elevated  above 
the  planes  of  the  angles,  making  equal  angles  with  their  sides, 
e£^ch  to  each,  viz.  the  angle  GAB  equal  to  the  angle  MDE,  and 
G/VC  to  MDF;  and  in  AG,  DM  let  any  points  G,  M  be  ta- 
ken, and  from  them  let  perpendiculars  GL,  MN  be  drawn  to 
the  planes  BAC,  EDF  meeting  these  planes  in  the  points  L,  Nj 
^nd  join  LA,  ND  ;  the  an^le  GAL  is  equal  to  the  angle  MDN? 

D 


Make  AH  equal  to  DM,  and  through  H  draw  HK  parallel 
to  GL:    but  GL  is  perpendicular  to  the  plane  BAC;    where- 
fore HK  is  perpendicular  »  to  the  same  plane  :    from  the  points  *  8.  H 
K,  N,  to  the  straight  lines  AB,  AC,   DE,   DF,  draw  perpen- 
Uicula/s   KB,   KC,   NE,   NF;    and  join   HB,  BC,   ME,  EF; 


244 


THE  ELEMENTS 


11. 

d  3. 

11. 


Book  XI.  because    HK   is   perpendicular   to   the   plane  BAG,  the  plane 
»— v—i^  HBK  which  passes  through  HK  is  at  right  angles  ^  to  the  plane 
bl8. 11.  BAG;  and  AB  is  drawn  in  the  plane  BAG  at  right  angles  to 
the  common  section  BK  of  the  two  planes ;    therefore   AB  is 
c  4.  def.  perpendicular  =   to  the   plane   HBK,  and  makes   right  angles  ^ 
with  every  straight  line  meeting  it  in  that  plane:  but  BH  meets 
^sf-  it  in  that  plane ;  therefore  ABH  is  a  right  angle  :  for  the  same 
reason,  DEM  is  a  right  angle,  and  is  therefore   equal    to  the 
angle  ABH:  and  the  angle  HAB  is  equal  to  the  angle  MDE. 
Therefore  in  the  two  triangles  HAB,  MDE  there  are  two  angles 
in   one    equal   to  two  angles   in   the  other,   each  to  each,   and 
one  side  equal  to  one  side,  opposite  to  one  of  the  equal  angles 
in  each,  viz.  HA  equal  to  DM ;  therefore  the  remaining  sides 
are  equal  *,  each  to  each :  wherefore  AB  is  equal  to  DE.      In 
the  same  manner,  if  HG  and  MF  be  joined,  it  may  be  demon- 
strated that  AG  is  equal  to  DF :   therefore,  since  AB  is  equal 
to  DE)  BA  and  AC  are  equal  to  ED  and  DF  ;  and  the  angle 


c  25. 1. 


BAG  is  equal  to  the  angle  EDF;  wherefore  the  base  BC  is 
f  4. 1.  equal  f  to  the  base  EF,  and  the  remaining  angles  to  the  remain- 
ing angles :  the  angle  ABG  is  therefore  equal  to  the  angle 
DliF:  and  the  right  angle  ABK  is  equal  to  the  right  angle 
DEN,  whence  the  remaining  angle  GBK  is  equal  to  the  re- 
maining angle  FEN :  for  the  same  reason,  the  angle  BGK  is 
equal  to  the  angle  EFN :  therefore  in  the  two  triangles  BGK, 
EFN,  there  are  two  angles  in  one  equal  to  two  angles  in  the 
other,  each  to  each,  and  one  side  equal  to  one  side  adjacent 
to  the  equal  angles  in  each,  viz.  BG  equal  to  EF ;  the  other 
sides,  tiierefore,  are  equal  to  the  other  sides;  BK  then  is  equal 
to  EN :  and  AB  is  equal  to  DE ;  wherefore  AB,  BK  are  equal 
'to  DE,  EN;  and  they  contain  right  angles;  wherefore  the 
base  AK  is  equal  to  the  base  DN :    and  since  AH  is  equal  to 


OF  EUCLID.  245 

DM,  the  square  of  AH  is  equal  to  the  square  of  DM:  but  theBookXI. 
squares  of  AK,  KH  are  equal  to  the  square  s  of  AH,  because  ^^-ymJ 
AKH  is  a  right  angle:  and  the  squares  of  DN,  NM  are  equal  g  47. 1. 
to  the  square  of  DM,  for  DNM  is  a  right  angle :  wherefore  the 
squares  of  AK,  KH  are  equal  to  the  squares  of  DN,  NM  ;  and  of 
those  the  square  of  AK  is  equal  to  the  square  of  DN  :  therefore 
the  remaining  square  of  KH  is  equal  to  the  remaining  square  of 
NM  ;  and  the  straight  line  KH  to  the  straight  line  NM :  and  be- 
cause HA,  AK  are  equal  to  MD,  DN  each  to  each,  and  the  base 
HK  to  the  base  MN  as  has  been  proved ;  therefore  the  angle  HAK 
is  equal  h  to  the  angle  MDN.     Q.  E.  D.  h  8. 1. 

Cor.  From  this  it  is  manifest,  that  if,  from  the  vertices  of  two 
equal  plane  angles,  there  be  elevated  two  equal  straight  lines 
containing  equal  angles  with  the  sides  of  the  angles,  each  to 
each  ;  the  perpendiculars  drawn  from  the  extremities  of  the 
equal  straight  lines  to  the  planes  of  the  first  angles  are  equal  to 
one  another. 


Another  Demonstration  of  the  Corollary, 

Let  the  plane  angles  B  AC,  EDF  be  equal  to  one  another,  and 
let  AH,  DM  be  two  equal  straight  lines  above  the  planes  of  the 
angles,  containing  equal  angles  with  BA,  AC  ;  ED,  DF,  each  to 
each,  viz.  the  angle  HAB  equal  to  MDE,  and  HAC  equal  to  the 
angle  MDF ;  a'nd  from  H,  M  let  HK,  MN  be  perpendiculars  to 
the  planes  BAC,  EDF :  HK  is  equal  to  MN. 

Because  the  solid  angle  at  A  is  contained  by  the  three  plane 
angles  BAC,  BAH,  HAC,  which  are,  each  to  each,  equal  to 
the  three  plane  angles  EDF,  EDM,  MDF  containing  the  solid 
angle  at  D  ;  the  solid  angles  at  A  and  D  are  equal :  and  therefore 
coincide  with  one  another  ;  to  wit,  if  the  plane  angle  BAC  be  ap- 
plied to  the  plane  angle  EDF,  the  straight  line  AH  coincides  with 
DM,  as  was  shown  in  prop.  B  of  this  book  :  and  because  AH  is 
equal  to  DM,  the  point  H  coincides  with  the  point  M  :  wherefore 
HK  which  is  perpendicular  to  the  plane  BAC  coincides  with 
MN  i  which  is  perpendicular  to  the  plane  EDF,  because  these  i  13. 11. 
planes  coincide  with  one  another :  therefore  HK  is  equal  to  MN. 
Q.  E.  D. 


246 
Book  XI. 


THE  ELEMENTS 


PROP.  XXXVI.    THEOR. 


See  N.  IF  three  straight  lines  be  proportionals,  the  solid 
parallelepiped  described  from  all  three  as  its  sides,  is 
equal  to  the  equilateral  parallelepiped  described  from 
the  mean  proportional,  one  of  the  solid  angles  of 
which  is  contained  by  three  plane  angles  equal,  each 
to  each,  to  the  three  plane  angles  containing  one  of 
the  solid  angles  of  the  other  figure. 

Let  A,  P,  C  be  three  proportionals,  viz.  A  to  B,  as  B  to  C. 
The  solid  described  from  A,  B,  C  is  equal  to  the  equilateral  so- 
lid described  from  B,  equiangular  to  the  other. 

Take  a  solid  angle  D  contained  by  three  plane  angles  EDF, 
FDG,  GDE;  and  make  each  of  the  straight  lines  ED,  DF, 
DG  equal  to  B,  and  complete  the  solid  parallelepiped  DH ; 
make  LK  equal  to  A,  and  at  the  point  K  in  the  straight  line 
a.  26. 11.  LK  make  a  a  solid  angle  contained  by  the  three  plane  angles 
LKM,  MKN,  NKL  equal   to  the  angles  EDF,  FDG,  GDE, 


O 


H 


M 


^ 


M^ 
^ 


E 


D 


B 


b  14.  6. 


rach  to  each  ;  and  make  KN  equal  to  B,  and  KM  equal  to 
C ;  and  complete  the  solid  parallelepiped  KO :  and  because,  a^ 
A  is  to  B,  so  is  B  to  C,  and  that  A  is  equal  to  LK,  and  B 
to  each  of  the  straight  lines  DE,  DF,  and  C  to  KM;  there- 
fore LK  is  to  ED,  as  DF  to  KM ;  that  is,  the  sides  about  the 
equal  angles  are  reciprocally  proportional;  therefore  the  pa- 
rallelogram LM  is  equal  •>  to  EF :  and  because  EDF,  LKM  are 
two  equal  plane  angles,  and  the  two  equal  straight  lines  DG, 
KN  are  drawn  from  their  vertices  above  their  planes,  and  con- 
tain equal  angles  with  their  sides;  therefore  the  perpendicu- 
>ars  from   the   points  G,  N,  to  the  planes  EDF,    LKM,    are 


OF  EUCLID. 


247 


equal <=  to  one  another:  therefore  the  solids  KO,  DH  are  of  the  Book XI. 
same  altitude ;  and  they  are  upon  equal  bases  LM,  EF ;  and  "^ — r— ' 
therefore  they  are  equal  ^  to  one  another:  but  the  solid  KO  is  c Cor. 35. 
described  from  the  three  straight  lines  A,  B,  C,  and  the  solid  DH  H- 
from  the  straight  line  B.  If,  therefore,  three  straight  lines,  £cc.  d  31. 11. 
Q.  E.  D. 


PROP.  XXXVII.    THEOR. 

IF  four  straight  lines  be  proportionals,  the  similar  See  n. 
solid  parallelepipeds  similarly  described  from  them 
shall  also  be  proportionals.  And  if  the  similar  paral- 
lelepipeds similarly  described  from  four  straight  lines 
be  proportionals,  the  straight  lines  shall  be  propor- 
tionals. 


Let  the  four  straight  lines  AB,  CD,  EF,  GH  be  proporti'^nals, 
viz.  as  AB  to  CD,  so  EF  to  GH ;  and  let  the  similar  parallelepi- 
peds AK,  CL,  EM,  GN  be  similarly  described  from  them.  AK 
is  to  CL,  as  EM  to  GN. 

Make  »  AB,  CD,  O,  P  continual  proportionals,  as  also  EF,  GH,  a  11.  6. 
Q,  R :  and  because  as  AB  is  to  CD,  so  EF  to  GH ;  and  that  CD 


\ 

^ 

\ 

\ 

fv^^ 


A 


\  \rAl 

B      C  I> 


M 

■^^-,^^^ 

\ 

^^ 

N 


^ 


E 


T 


R 


F  G        H  Q 

isbto  O,  as  GH  to  Q,  and  O  to  P,  as  Q  to  R ;    therefore,  e^b  11.5. 
a(/uali  c,  AB  is  to  P,  as  EF  to  R  :  but  as  AB  to  P,  so  d  is  the  solid  c  22.  5. 
AK  to  the  solid  CL ;  and  as  EF  to  R,  so  ^  is  the  solid  EM  to  the 
solid  GN :  therefore  •>  as  the  solid  AK  to  the  solid  CL,  so  is  the  dCor.JcJ. 
solid  EM  to  the  solid  GN. 


248 


THE  ELEMENTS 


Book  XI.  But  let  the  solid  AK  he  to  the  solid  CL,  as  the  solid  EM  to  the 
»— v^-'  solid  GN :  the  straight  line  AB  is  to  CD,  as  EF  to  GH. 
e  27. 11.  Take  AB  to  CD,  as  EF  to  ST,  and  from  ST  describe  «  a  solid 
parallelepiped  SV  similar  and  similarly  situated  to  either  of  the 
solids  EM,  GN:  and  because  AB  is  to  CD,  as  EF  to  ST  ;  and 
that  from  AB,  CD  the  solid  parallelepipeds  AK,  CL  are  similarly 
described;  and,  in  like  manner,  the  solids  EM,  SV  from  the 
straight  lines  EF,  S  L  ;   therefore  AK  is  to  CL,  as  EM  to  SV : 


f9.  5. 


\ 

Nf 

V 

\1 

\ 


A. 


3     C 


o 


-V 


N 


fx 


H 


Q      R 


but,  by  the  hypothesis,  AK  is  to  CL,  as  EM  to  GN :  therefore 
GN  is  equal  f  to  SV  :  but  it  is  likewise  similar  and  similarly  situ- 
ated to  SV ;  therefore  the  planes  which  contain  the  solids  GN, 
SV  are  similar  and  equal,  and  their  homologous  sides  GH,  ST 
equal  to  one  another :  and  because  as  AB  to  CD,  so  EF  to  ST, 
and  that  ST  is  equal  to  GH  ;  AB  is  to  CD,  as  EF  to  GH.  There- 
fore, if  four  straight  linos.  Sec.     Q.  E.  D. 


PROP.  XXXVHL    THEOR. 


SeeN. 


"  IF  a  plane  be  perpendicular  to  another  plane,  and 
"  a  straight  line  be  drawn  from  a  point  in  one  of  the 
"  planes  perpendicular  to  the  other  plane,  this  straight 
*'  line  shall  fall  on  the  common  section  of  the  planes." 

"  Let  the  plane  CD  be  perpendicular  to  the  plane  AB,  and  let 
«  AD  be  their  common  section  ;  if  any  point  E  be  taken  in  the 
"  plane  CD,  the  perpendicular  drawn  from  E  to  the  plane  AB 
« shall  fall  on  AD.  -  / 


OV  EUCLID. 


249 


11. 


'•■  For,  if  it  does  not,  let  it,  if  possible,  fall  elsewhere,  as  EF ;  Book  XL 
•••  and  let  it  meet  the  plane  AB  in  the  point  F  ;  and  from  F  draw  »,  ^-— v— ' 
"  in  the  plane  AB,  a  perpendicular  FG  to  DA,  which  is  also  per-  a  12. 1. 
"  pendicular''  to  the  plane  CD  ;  and  join  EG:  then  because  FG  b  4.  def. 
"  is  perpendicular   to  the    plane 
"  CD,  and  the  straight  line  EG,      ( 
"  which  is  in  that  plane,  meets  it ; 
"  therefore  FGE  is  a  right  an- 
"  gle«:  but   EF  is  also  at  right 
"  angles  to  the  plane  AB ;   and 
"  therefore  EFG  is  a  right  angle  :      u 
"  wherefore  two  of  the  angles  of 
"  the  triangle  EFG  are  equal  to- 
"  gether    to   two    right    angles : 

"  which  is  absurd  :  tnerefore  the  perpendicular  from  the  point 
"  E  to  the  plane  AB,  does  not  fall  elsewhere  than  upon  the 
"  straight  line  AD  :  it  therefore  falls  upon  it.  If  therefore  a 
«  plane,"  S^c.     Q.  E.  D. 


PROP.  XXXIX.    THEOR. 


IN  a  solid  parallelepiped,  if  the  sides  of  two  of  the  See  N. 
opposite  planes  be  divided  each  into  two  equal  parts, 
the  common  section  of  the  planes  passing  through  the 
points  of  division,  and  the  diameter  of  the  solid  paral- 
lelepiped  cut  each  other  into  two  equal  parts. 


Let  the  sides  of 
the  opposite  planes 
CF,  AH  of  the  solid 
parallelepiped  AF, 
be  divided  each  into 
two  equal  parts  in 
the  points  K,  L,  M, 
N;  X,  O,  P,  R; 
and  join  KL,  MN, 
XO,  PR  :  and  be- 
cause  DK,  CL  are 
equal  and  parallel, 
KL  is  parallel  *  to 
DC :  for  the  same 
reason,  MN  is  pa- 
rallel to  BA :  and 
BA  is   parallel   to 


»  33. 1, 


2  I 


250 

Book  XI 
b  9.  11. 


c  29.  1. 


d4.  1. 


e  li  1. 


9,  33.  !• 


f  15  1. 


6  26.1. 


THE  ELEMENTS 

DC  ;  therefore,  because  KL,  BA  are  each  of  them  parallel  to 
DC,  and  not  in  the  same  plane  with  it,  KL  is  parallel  *>  to  BA : 
and  because  KL,  MN  are  each  of  them  parallel  to  BA,  and  not 
in  the  same  plane  with  it,  KL  is  parallel^  to  MN  ;  wherefore 
KL,  MN  are  in  one  plane.  In  like  manner,  it  may  be  proved, 
that  XO,  PR  are  in  one  plane.  Let  YS  be  the  common  section 
of  the  planes  KN,  XR  ;  and  DG  the  diameter  of  the  solid  paral- 
lelepiped AF  :  YS  and  DG  do  meet,  and  cut  one  another  into  two 
equal  parts. 

Join  DY,  YE,  BS,  SG.  Because  DX  is  parallel  to  OE,  the 
alternate  ano;!es  DXY,  YOE  are  equal*:  to  one  another:  and  be- 
cause DX  is  equal 

to  OE,  and  XY  to     D  K  F 

YO,     and     contain 
equal     angles,    the 
base  DY  is   equal  ^ 
to  the  base  YE,  and 
the  other  angles  are 
equal ;  therefore  the 
angle  XYD  is  equal 
to  the  angle  OYE, 
and      DYE      is      a 
straight  f^   line  :    foi* 
the     same      reason 
BSG    is   a  straight  B 
line,  and  BS  equal 
to  SG  :  and  because 
CAis  equal  and  pa- 
rallel to  DB,  and  al- 
so ecjual  and  paral- 
lel to  EG  ;  therefore  DB  is  equal  and  parallel''  to  EG  :  and  DE, 
BG  join  their  extremities;  therefore  DE  is  equal  and  parallel » 
to  BG :  and  DG,  YS  are  drawn  from  points  in  the  one  to  points 
in  the  other ;  and  are  therefore  in  one  plane :  whence  it  is  mani- 
fest, that  DG,  YS  must  meet  one  another ;  let  them  meet  in  T ; 
and  because  DE  is  parallel  to  BG,  the  alternate  angles  EDT, 
BGT  are  equal «;  and  the  angle  DTY  is  equal  ^  to  the  angle 
GTS:  therefore  in  the  triangles  DTY,  GTS  there  are  two  an- 
ples  in  the  one  equal  to  two  angles  in  the  other,  and  one  side 
equal  to  one  side,  opposite  to  two  of  the  equal  angles,  viz.  DY 
to  GS;  for  they  are  the  halves  of  DE,  BG :  therefore  the  re- 
maining  sides   are    equals,   each  to  each.      Wherefore   DT  is 
equal  to  TG,  and  YT  equal  to  TS.    Vv^herefore,  if  in  a  solid,  8cc. 
^'  E-  I^' 


OF  EUCLID. 


PROP.  XL.    THEOR. 


IF  there  be  two  triangular  prisms  of  the  same  al- 
titude, the  base  of  one  of  which  is  a  parallelogram, 
and  the  base  of  the  other  a  triangle ;  if  the  parallelo- 
gram be  double  of  the  triangle,  the  prisms  shall  be 
equal  to  one  another. 

Let  the  prisms  ABCDEF,  GHKLMN  be  of  the  same  altitude, 
the  first  whereof  is  contnined  by  the  two  triangles  ABE,  CDF 
and  the  three  parallelograms  AD,  DE,  EC;  and  the  other  by 
the  two  triangles  GHK,  LMN  and  the  three  parallelograms  LHj 
HN,  NG  ;  and  let  one  of  them  have  a  parallelogram  AF,  and  the 
other  a  triangle  GHK  for  its  base  ;  if  the  parallelogram  AF  be 
double  of  the  triangle  GHK,  the  prism  ABCDEF  is  equal  to  the 
prism  GHKLMN. 

Complete  the  solids  AX,  GO ;  and  because  the  parallelogram 
AF  is  double  of  the  triangle  GHK ;   and  the  parallelogram  HK 


double  a  of  the  same  triangle  ;    therefore  the  parallelogram  AFa  34. 1. 

is  equal  to  HK.     But  solid  parallelepipeds  upon  equal  bases,  and 

of  the  same  altitude,  are  equal  ^  to  one  another.     Therefore  theb31.il. 

solid  AX  is  equal  to  the  solid  GO  ;    and  the  prism  ABCDEF  is 

half  c  of  the  solid  AX;    and  the  prism  GHKLMN  half  <=  of  the  c  28. 11. 

solid  GO.     Therefore  the  prism  ABCDEF  is  equal  to  the  prism 

GHKLMN.     Wherefore,  if  there  be  two,  &c.     Q.  E.  D. 


THE 


ELEMENTS  OF  EUCLID. 


BOOK  XII. 


LEMMA  I. 

B.  XII.  Which  is  the  first  proposition  of  the  tenth  book,  and  is  necessary 
*— -v"*^       to  some  of  the  propositions  of  this  book. 


See  N.  IF  from  the  greater  of  two  unequal  magnitudes 
there  be  taken  more  than  its  half,  and  from  the  re- 
mainder more  than  its  half,  and  so  on,  there  shall  at 
length  remain  a  magnitude  less  than  the  least  of  the 
proposed  magnitudes. 


D 


K— 


H— 


Let  AB  and  C  be  two  unequal  magnitudes,  of  which  AB  is 
the  greater.      If  from  AB  there  be  taken  more 
than  its  half,  and  from  the  remainder  more  than 
its  half,  and  so  on,  there  shall  at  length  remain  a 
magnitude  less  than  C. 

For  C  may  be  multiplied  so  as  at  length  to 
become  greater  than  AB.  Let  it  be  so  multi- 
plied, and  let  DE  its  multiple  be  greater  than 
AB,  and  let  DE  be  divided  into  DF,  FG,  GE, 
each  equal  to  C.  From  AB  take  BH  greater 
than  its  half,  and  from  the  remainder  AH 
take  HK  greater  than  its  half,  and  so  on,  until 
there  be  as  many  divisions  in  AB  as  there  are 
in  DE:  and  let  the  divisions  in  AB  be  AK, 
KII,  HB ;  and  the  divisions  in  DE  be  DF,  FG, 
GE.     And  because  DE  is  greater  than  AB,  and        B    C     K 


F' 


G— 


THE  ELEMENTS,  See.  252 

that  EG  taken  from  DE  is  not  greater  than  its  half,  but  BH  ta-  B.  XII. 
ken  from  AB  is  greater  than  its  half;  therefore  the  remainder  ^— -y^*^ 
GD  is  greater  than  the  remainder  HA.  Again,  because  GD  is 
greater  than  HA,  and  that  GF  is  not  greater  than  the  half  of 
GD,  but  HK  is  greater  than  the  half  of  HA ;  therefore  the  re- 
mainder FD  is  greater  than  the  remainder  AK.  And  FD  is 
equal  to  C,  therefore  C  is  greater  than  AK  ;  that  is,  AK  is  less 
than  C.     Q.  E.  D. 

And  if  only  the  halves  be  taken  away,  the  same  thing  may  in 
the  same  way  be  demonstrated. 


PROP.  I.  THEOR. 


SIMILAR  polygons  inscribed  in  circles  are  JLo  one 
another  as  the  squares  of  their  diameters. 

Let  ABCDE,  FGHKL  be  two  circles,  and  in  them  the  si- 
milar polygons  ABCDE,  FGHKL  ;  and  let  BM,  GN  be  the 
diameters  of  the  circles ;  as  the  square  of  BM  is  to  the  square 
of  GN,  so  is  the  polygon  ABCDE  to  the  polygon  FGHKL. 

Join  BE,  AM,  GL,  FN  :  and  because  the  polygon  ABCDE  is 
similar  to  the  polygon  FGHKL,  and  similar  polygons  are  divided 
into  similar  triangles;  the  triangles  ABE,  FGL  are  similar  and 
A  F 


D 

equiangular'' ;  and  therefore  the  angle  AEB  is  equal  to  the  angle  b  6.  6, 
FLG  :  but  AEB   is  equal  <=  to  AMB,  because   they   stand  up- c  21. 3, 
on  the  same  circumference  ;  and  the  angle  FLG  is,  for  the  same 
reason,  equal  to  the  angle  FNG  ;  therefore  also  the  angle  AMB 
is   equal   to  FNG ;    and  the  right  angle  BAM  is  equal  to  the 
right  d  angle  GFN  ;  wherefore  the  remaining  angles  in  the  tri-d  51- 3, 
angles  ABM,  FGN  are  equal,  and  they  are  equiangulr.r  to  one 


254 


THE  ELEMENTS 


B.  XII.  another  :  therefore  as  BM  to  GN,  so  e  is  BA  to  GF  ;  and  there- 
^-"->r— -^  fore  the  duplicate  ratio  of  BM  to  GN  is  the  same  ^  with  the  du- 
e  4  6.  plicate  ratio  of  B  A  to  GF  :  but  the  ratio  of  the  square  of  BM  to 
f  10.  def.  the  square  of  GN,  is  the  duplicate  e  ratio  of  that  which  BM  has 
5.  &  22.5.  to  GN  ;  and  the  ratio  of  the  polys^on  ABODE  to  the  polygon 
g20.  6.     FGHKL  is  the  duplicate  sof  that  which  BA  has  to  GF  :  there- 


fore as  the  square  of  BM  to  the  square  of  GN,  so  is  the  polygon 
ABCDE  to  the  polygon  FGHKL.  Wherefore  similar  polygons, 
&c.    Q.  E.  D. 


PROP.  n.    THEOR. 


See  N. 


CIRCLES  are  to  one  another  as  the  squares  of 
their  dicimeters. 


Let  A  BCD,  EFGH  be  two  circles,  and  BD,  FH  their  diame- 
ters :  as  the  square  of  BD  to  the  square  of  FH,  so  is  the  circle 
ABCD  to  the  circle  EFGH. 

For,  if  it  be  not  so,  the  square  of  BD  shall  be  to  the  square 
of  FH,  as  the  circle  ABCD  is  to  some  space  either  less  than  the 
circle  EFGH,  or  greater  than  it*.  First,  let  it  be  to  a  space 
S  less  than  the  circle  EFGH  ;  and  in  the  circle  EFGH 
describe  the  square  EFGH  :  this  square  is  greater  than 
half  of  the  circle  EFGH  ;  because  if,  through  the  points 
E,    F,    G,    H,    there    be    drawn    tangents    to    the    circle,    the 

•  For  there  is  some  square  equal  to  the  circle  ABCD  :  let  P  be  the  side  of 
it,  and  to  three  straight  lines  BD,  FH,-  and  P,  there  can  be  a  fourth  propor- 
tional ;  let  this  be  Qj  therefore  the  squares  of  these  four  straight  lines  are 
proportionals  :  that  is,  to  the  squares  of  BD,  FH  and  the  circle  ABCD,  it  is 
possible  there  may  be  a  fourth  proportional.  Let  this  be  S.  And  in  like  man- 
ner are  to  be  understood  .some  things  in  some  of  the  following  propositions. 


OF  EUCLID. 


255 


square  EFGH  is  half*  of  the  square  described  about  the  circle  ;  B.  XII. 
and  the  circle  is  less  than  the  square  described  about  it;  there-  ^-^-v-i-^ 
fore  the  square  EFGH  is  greater  than  half  of  the  circle.  Divide  a  41. 1. 
the  circumferences  EF,  FG,  GH,  HE  each  into  two  equal  parts 
in  the  points  K,  L,  M,  N,  and  join  EK,  KF,  FL,  LG,  GM,  WH, 
HN,  NE:  therefore  each  of  the  triangles  EKF,  FLG,  GMH, 
HNE  is  greater  than  half  of  the  segment  of  the  circle  it  stands 
in  ;  because,  if  straight  lines  touching  the  circle  be  drawn  through 
the  points  K,  L,  M,  N,  and  parallelograms  upon  the  straight 
lines  EF,  FG,  GH,  HE  be  completed  ;  each  of  the  triangles  EKF, 
FLG,  GMH,  HNE  shall  be  the  half  a  of  the  parallelogram  in 
which  it  is:  but  every  segment  is  less  than  the  parallelogram  in 
which  it  is:  wherefore  each  of  the  triangles  EKF,  FLG,  GMH, 
HNE  is  greater  than  half  the  segment  of  the  circle  which  con- 
tains it:  and  if  these  circumferences  before  named  be  divided 
each  into  two  equal  parts,  and  their  extremities  be  joined  by 
straight  lines,  by  continuing  to  do  this,  there  will  at  length  re- 


main segments  of  the  circle  which,  together,  shall  be  less  than 
the  excess  of  the  circle  EFGH  above  the  space  S:  because,  by 
the  preceding  lemma,  if  from  the  greater  of  two  unequal  raag^ 
nitudes  there  be  taken  more  than  its  half,  and  from  the  rem  in- 
der  more  than  its  half,  and  so  on,  there  shall  at  length  remair.  a 
magnitude  less  than  the  least  of  the  proposed  magnitudes.  Let 
then  the  segments  EK,  KF,  FL,  LG,  GM,  MH,  HN,  NF  be 
those  that  remain  and  are  together  less  tl)an  the  excess  of  the 
circle  EFGH  above  S:  therefore  the  rest  of  the  circle,  viz.  the 
polygon  EKFLGMHN,  is  greater  than  the  space  S.  Describe 
likewise  in  the  circle  AliCD  the  polygon  AXBOCPDR  similar 
to  the  polygon  EKFLGMHN :  as,  therefore,  the  square  of  BD 
is  to  the  square  of  FH,  so  ^  is  the  polygon  AXBOCPDR  to  the  b  1. 12: 
polygon  EKFLGMHN:  but  the  square  of  BD  is  also  to  the 
square  of  FII,  a-,  the  circle  ABCD  i§  to  the  space  S:    thersfove 


*56 


THE  ELEMENTS 


B.  Xll.  as  the  circle  ABCD  is  to  the  space  S,  so  is  *=  the  polygon 
AXBOCPDR  to  the  polygon  EKFLGMHN:  but  the  circle 
ABCD  is  greater  than  the  polygon  contained  in  it :  wherefore, 
the  space  S  is  greater ^  than  the  polygon  EKFLGMHN:  but  it 
is  likewise  less,  as  has  been  demonstrated ;  which  is  impossible. 
Therefore  the  square  of  BD  is  not  to  the  square  of  FH  as  the 
circle  ABCD  is  to  any  space  less  than  the  circle  EFGH.  In  the 
same  manner  jt  may  be  demonstrated,  that  neither  is  the  square 
of  FH  to  the  square  of  BD,  as  the  circle  EFGH  is  to  any  space 
less  than  the  circle  ABCD.  Nor  is  the  square  of  BD  to  the 
square  of  FH,  as  the  circle  ABCD  is  to  any  space  greater  than 
the  circle  EFGH  :  for,  if  possible,  let  it  be  so  to  T,-a  space  great^ 
er  than  the  circle  EFGH :  therefore,  inversely,  as  the  square  of 
FH  to  the  square  of  BD,  so  is  the  space  T  to  the  circle  ABCD. 


But  as  the  space  f  T  is  to  the  circle  ABCD,  so  is  the  circle 
EFGH  to  some  space,  which  must  be  less  «*  than  the  circle 
ABCD,  because  the  space  T  is  greater,  by  hypothesis,  than  the 
circle  EFGH.     Therefore,  as  the  square  of  FH  is  to  the  square 


t  tot,  as  in  the  foregoing-  note,  at  *,  it  was  explained  how  it  was  possible 
there  could  be  a  fourth  proportional  to  the  squares  of  BD,  FH,  and  the  circle 
ABCD,  which  was  named  S.  So,  in  like  manner,  there  can  be  a  fourth  pro- 
portional to  this  other  space,  named  T,  and  the  circles  ABCD,  EFGH.  And 
the  like  is  t^  be  understood  in  some  of  the  following  propositions. 


OF  EUCLID. 


257 


of  BD,  so  is  the  circle  EFGH  to  a  space  less  than  the  circle  B.  XII. 
ABCD,  which  has  been  demonstrated  to  be  impossible :  there-  ^— y-^J 
fore  the  square  of  BD  is  not  to  the  square  of  FH  as  the  circle 
ABCD  is  to  any  space  greater  than  the  cii'cle  EFGH:  and  it  has 
been  demonstrated,  that  neither  is  the  square  of  BD  to  the  square 
of  FH,  as  the  circle  ABCD  to  any  space  less  than  the  circle 
EFGH :  wherefore,  as  the  square  of  BD  to  the  square  of  FH, 
so  is  the  circle  ABCD  to  the  circle  EFGH^.  Circles,  therefore, 
are,  kc.    Q.  E.  D. 


PROP.  III.    THEOR. 


EVERY  pyramid  having  a  triangular  base  may  be  See  n. 
divided  into  two  equal  and  similar  pyramids  having 
triangular  bases,  and  which  are  similar  to  the  whole 
pyramid,  and  into  two  equal  prisms  which  together 
are  greater  than  half  of  the  whole  pyramid. 


Let  there  be  a  pyramid  of  which  the  base  is  the  triangle  ABC, 
and  its  vertex  the  point  D :  the  pyramid  ABCD  may  be  divided 
into  two  equal  and  similar  pyramids  having 
triangular  bases,  and  similar  to  the  whole ; 
and  into  two  equal  prisms  which  together 
af-e  greater  than  half  of  the  whole  pyramid. 

Divide  AB,  BC,  CA,  AD,  DB,  DC,  each 
into  two  equal  parts  in  the  points  E,  F,  G, 
H,  K,  L,  and  join  FH,  EG,  GH,  HK,  KL, 
LH,  EK,  KF,  FG.  Because  AE  is  equal  to 
EB,  and  AH  to  HD,  HE  is  parallel » to  DB: 
for  the  same  reason,  HK  is  parallel  to  AB : 
therefore  HEBK  is  a  parallelogram,  and  HK 
equal  ^  to  EB  :  but  EB  is  equal  to  AE  ;  there- 
fore also  AE  is  equal  to  HK :  and  AH  is  equal 
to  HD  ;  wherefore  EA,  AH  are  equal  to  KH, 
HD,  each  to  each ;  and  the  angle  EAH  is 
equal  c  to  the  angle  KHD  ;  therefore  the  base 
EH  is  equal  to  the  base  KD,  and  the  triangle  B 


a  2.  6. 


b  34.  1. 


c29.1. 


:j:  Because  as  a  fourth  proportional  to  the  squares  of  BD,  FH  and  the  circle 
ABCD  is  possible,  and  that  it  can  neither  be  less  nor  greater  than  the  circle 
F'FGH,  it  must  be  equal  to  it.  if 

2  K 


25S 


THE  ELEMENTS 


B.  XII.  AEH  equal  <*  and  similar  to  the  triangle  HKD :  for  the  same 
reason,  the  triangle  AGH  is  equal  and  similar  to  the  triangle 
HLD  :  and  because  the  two  straight  lines  EH,  HG  which  meet 
one  another  are  parallel  to  KD,  DL  that  meet  one  another,  and 
are  not  in  the  same  plane  with  them,  they  contain  equals  angles; 
therefore  the  angle  EHG  is  equal  to  the  angle  KDL.  Again, 
because  EH,  HG  are  equal  to  KD,  DL,  each  to  each,  and  the 
angle  EHG  equal  to  the  angle  KDL;  therefore  the  base  EG  is 
equal  to  the  base  KL  :  and  the  triangle  EHG  equal  ^  and  similar 
to  the  triangle  KDL  :  for  the  same  reason,  the  triangle  AEG  is 
also  equal  and  similar  to  the  triangle  HKL.  Therefore  the  py- 
ramid of  which  the  base  is  the  triangle  AEG,  and  of  which  the 
f  C.  II.  vertex  is  the  point  H,  is  equal  f  and  similar  to  the  pyramid  the 
base  of  which  is  the  triangle  HKL,  and  ver- 
tex the  point  D  :  and  because  HK  is  parallel 
to  AB  a  side  of  the  triangle  ADB,  the  tri- 
angle ADB  is  equiangular  to  the  triangle 
g  4.  6.  HDK,  and  their  sides  are  proportionals  e  : 
therefore  the  triangle  ADB  is  similar  to 
the  triangle  HDK :  and  for  the  same  rea- 
son, the  triangle  DBG  is  similar  to  the  tri- 
angle DKL ;  and  the  triangle  ADC  to  the 
triangle  HDL ;  and  also  the  triangle  ABC 
to  the  triangle  AEG  :  but  the  triangle  AEG 
is  similar  to  the  triangle  HKL,  as  before 
was  proved  ;  therefore  the  triangle  ABC  is 
b  21.  6.  similar  ^  to  the  triangle  HKL.  And  the 
pyramid  of  which  the  base  is  the  triangle 
ABC,  and  vertex  the  point  D,  is  therefore 
i  B.  11.  &  similar  » to  the  pyramid  of  which  the  base 

Jl.  def.     is  the  triangle  HKL,  and  vertex  the  same  point  D  :    but  the  py- 
11-  ramid  of  which  the  base  is  the  triangle  HKL,  and  vertex  the 

point  D,  is  similar,  as  has  been  proved,  to  the  pyramid  the  base 
of  which  is  the  triaui^le  AEG,  and  vertex  the  point  H:  where- 
fore the  pyramid  the  base  of  which  is  the  triangle  ABC,  and  ver- 
tex the  point  D,  is  similar  to  the  pyramid  of  which  the  base  is  the 
triangle  AEG,  and  vertex  H :  therefore  each  of  the  pyramids 
AEGH,  HKLD  is  similar  to  the  whole  pyramid  ABCD  :  and  be- 
k  41. 1  cause  BE  is  equal  to  EC,  the  parallelogram  EBEG  is  double  ^  of 
the  triangle  GEC:  but  when  there  are  two  prisms  of  the  same  alti- 
tude, of  which  one  has  a  parallelogram  for  its  base,  and  the  other 
a  triangle  that  is  half  of  the  parallelogram,  these  prisms  are  equal 
a  40.  11.  *  to  one  another;  thei'efore  the  prism  having  the  parallelogram 
EBIG  for  its  base,  and  the  straight  line  KH  opposite  to  it,  is 
equal  to  the  prism  having  the  triangle  GFC  for  its  base,  and 
tVse  triangle  HKL  opposite  to  it ;    for  they  are  of  the  same  aiti- 


OF  EUCLID.  559 

tude,  because  they  are  between  the  parallel  l'  planes  ABC,  HKL:  B.  Xlt. 
and  it  is  manifest  that  each  of  these  prisms  is  greater  than  either  ^'-r-i«i 
of  the  pyramids  of  which  the  triangles  AEG,  HKL  are  the  b  15. 11. 
bases,  and  the  vertices  the  points  H,  D ;  because,  if  EF  be  join- 
ed, the  prism  having  the  parallelogram  EBFG  for  its  base,  and 
KH  the  straight  line  opposite  to  it,  is  greater  than  the  pyramid 
of  which  the  base  is  the  triangle  EBF,  and  vertex  the  point  K ; 
but  this  pyramid  is  equal  ^  to  the  pyramid  the  base  of  which  is  C  C.  It 
the  triangle  AEG,  and  vertex  the  point  H ;  because  they  are 
contained  by  equal  and  similar  planes :  wherefore  the  prism 
having  the  parallelogram  EBi^^G  for  its  base,  and  opposite  side 
KH,  is  greater  than  the  pyramid  of  which  the  base  is  the  trian- 
gle AEG,  and  vertex  the  point  H :  and  the  prism  of  which  the 
base  is  the  parallelogram  EBFG,  and  opposite  side  KH,  is  equal 
to  the  prism  having  the  triangle  GFC  for  its  base,  and  HKL  the 
triangle  opposite  to  it:  and  the  pyramid  of  which  the  base  is  the 
triangle  AEG,  and  vertex  H,  is  equal  to  the  pyramid  of  which 
the  base  is  the  triangle  HKL,  and  vertex  D :  therefore  the  two 
prisms  before  mentioned  are  greater  than  the  two  pyramids  of 
which  the  bases  are  the  triangles  AEG,  HKL,  and  vertices  the 
points  H,  D.  Therefore  the  whole  pyramid  of  which  the  base 
is  the  triangle  ABC,  and  vertex  the  point  D,  is  divided  into  two 
equal  pyramids  similar  to  one  another,  and  to  the  whole  pyra- 
mid ;  and  into  two  equal  prisms;  and  the  two  prisms  are  toge» 
ther  greater  than  half  of  the  whole  pyramid.    Q.  E.  D. 


THE  ELEMENTS 


PROP.  IV.    THEOR. 


See  Jf.  IF  there  be  two  pyramids  of  the  same  altitude,  up- 
on triangular  bases,  and  each  of  them  be  divided  in- 
to two  equal  pyramids  similar  to  the  whole  pyramid, 
and  also  into  two  equal  prisms ;  and  if  each  of  these 
pyramids  be  divided  in  the  same  manner  as  the  first 
two,  and  so  on ;  as  the  base  of  one  of  the  first  two 
pyramids  is  to  the  base  of  the  other,  so  shall  all  the 
prisms  in  one  of  them  be  to  all  the  prisms  in-the  other, 
that  are  produced  by  the  same  number  of  divisions. 

Let  there  be  two  pyramids  of  the  same  altitude  upon  the  tri- 
angular bases  ABC,  DEF,  and  having  their  vertices  in  the 
points  G,  H  ;  and  let  each  of  them  be  divided  into  two  equal 
pyramids  similar  to  the  whole,  and  into  two  equal  prisms  ;  and 
let  each  of  the  pyramids  thus  made  be  conceived  to  be  divided 
in  the  like  manner,  and  so  on ;  as  the  base  ABC  is  to  the  base 
DEF,  so  are  all  the  prisms  in  the  pyramid  ABCG  to  all  the 
prisms  in  the  pyramid  DEFH  made  by  the  same  number  of  di- 
visions. 

Make  the  same  construction  as  in  the  foi'egoing  proposition : 
and  because  BX  is  equal  to  XC,  and  AL  to  LC,  therefore  XL 

a  2. 6.  is  parallel  »  to  AB,  and  the  triangle  ABC  similar  to  the  tri- 
angle LXC :  for  the  same  reason,  the  triangle  DEF  is  similar 
to  RVF :  and  because  BC  is  double  of  CX,  and  EF  double  of 
FV,  therefore  BC  is  to  CX,  as  EF  to  FV  :  and  upon  BC,  CX 
are  described  the  similar  and  similarly  situated  rectilineal 
figures  ABC,  LXC ;  and  upon  EF,  FV,  in  like  manner,  are 
described  the  similar  figures  DEF,  RVF :    therefore,  as  the  tri- 

b  22.  6.  angle  ABC  is  to  the  triangle  LXC,  so  ^  is  the  triangle  DEF  to 
the  triangle  RVF,  and,  by  permutation,  as  the  triangle  ABC 
to  the  triangle  DEF,  so  is  the  triangle  LXC  to  the  triangle 
RV^F :   and  because  the  planes  ABC,  OMN,  as  also  the  planes 

cl5. 11.  DEF,  STY,  are  parallel  <=,  the  perpendiculars  drawn  from  the 
points  G,  H  to  the  bases  ABC,  DEF,  which,  by  the  hypothe- 
sis, are  equal  to  one   another,  shall  be  cut  each  into  two  equal 

d  17. 11.  ^  parts  by  the  planes  OMN,  STY,  because  the  straight  lines 
GC,  HF  are  cut  into  two  equal  parts  in  the  points  N,  Y  by 
the  same  planes:  therefore  the  prisms  LXCOMN,  RVFSTY 
are  of  the  same  altitude ;    and,  therefore,  as  the  base  LXC  to 


OF  EUCLID. 


261 


the  base  RVF;   that  is,  as  the  triangle  ABC  to  the  triangle  B.  XII. 
DEF,  so  a  is  the  prism  having  the  triangle  LXC  for  its  base,  ^  — v— -^ 
and  OMN  the  triangle  opposite  to  it,  to  the  prism  of  which  the  a  Cor.  32- 
base  is  the  triangle  RVF,  and  the  opposite  triangle  STY:  and   ^^• 
because   the  two  prisms  in  the  pyramid  ABCG  are   equal   to 
one   another,  and  also  the  two  prisms  in  the  pyramid  DEFH 
equal  to  one  another,  as  the  prism  of  which  the  base  is  the  pa- 
rallelogram KBXL  and  opposite  side  MO,  to  the  prism  having 
the  triangle  LXC  for  its  base,  and  OMN  the  triangle  opposite 
to  it,  so  is  the  prism  of  which  the  base  •>  is  the  parallelogram  b  7.  5. 
PEVR,  and  opposite  side  TS,  to  the  prism  of  which  the  base 
is  the  triangle  RVF,  and  opposite  triangle  STY.      Therefore, 
componendo,  as  the  prisms   KBXLMO,   LXCOMN   together 


G 


H 


are  unto  the  prism  LXCOMN,  so  are  the  prisms  PEVRTS, 
RVFSTY  to  the  prism  RVFSTY;  and,  permutando,  as  the 
prisms  KBXLMO,  LXCOMN  are  to  the  prisms  PEVRTS, 
RVFSTY,  so  is  the  prism  LXCOMN  to  the  prism  RVFSTY: 
but  as  the  prism  LXCOMN  to  the  prism  RVFSTY.  so  is,  as 
has  been  proved,  the  base  ABC  to  the  base  DEF:  therefore, 
as  the  base  ABC  to  the  base  DEF,  so  are  the  two  prisms  in  the 
pyramid  ABCG  to  the  two  prisms  in  the  pyramid  DEFH :  and 
likeAvise  if  the  pyramids  now  made,  for  example,  the  two  OMNG, 
STYH,  be  divided  in  the  same  manner ;  as  the  base  OMN  is  to 
the  base  STY,  so  shallthe  two  prisms  in  the  pyramid  OMNG 
be  to  the  two  prisms  in  the  pyramid  STYH :  but  the  base  OMN 
is  to  the  base  STY,  as  the  base  ABC  to  the  base  DEF ;  there- 
fore: as  the  base  ABC  to  the  base  DEF,  so  are  the  two  prisms 


263  THE  ELEMENTS 

B'.  XII.  in  the  pyramid  ABCG  to  the  two  prisms  in  the  pyramid  DF<FH ; 

^■— y~-^  and  so  are  the  two  prisms  in  the  pyramid  OMNG  to  the  two 
prisms  in  the  pyramid  STYH  ;  and  so  are  all  four  to  all  four: 
and  the  same  thing  may  be  shown  of  the  prisms  made  by  divid- 
ing the  pyramids  AKLO  and  DPRS,  and  of  all  made  by  the 
same  number  of  divisions.     Q.  E.  D. 


PROP.  V.     THEOR. 

SeeN.  PYRAMIDS  of  the  same  altitude  which  have  tri- 
angular bases,  are  to  one  another  as  their  bases. 

Let  the  pyramid's  of  which  the  triangles  ABC,  DEF  are  the 
bases,  and  of  which  the  vertices  are  the  points  G,  H,  be  of  the 
same  altitude :  as  the  base  ABC  to  the  base  DEF,  so  is  the  py- 
ramid ABCG  to  the  pyramid  DEFH. 

For,  if  it  be  not  so,  the  base  ABC  must  be  to  the  base  DEF, 
as  the  pyramid  ABCG  to  a  solid  either  less  than  the  pyramid 
DEFH,  or  greater  than  it*.  First,  let  it  be  to  a  solid  less  than 
it,  viz.  to  the  solid  Q :  and  divide  the  pyramid  DEFH  into 
two  equal  pyramids,  similar  to  the  whole,  and  into  two  equal 

a  3. 12.  pi'isms :  therefore  these  two  prisms  are  greater  ^  than  the  half 
of  the  whole  pyramid.  And  again,  let  the  pyramids  made  by 
this  division  be  in  like  manner  divided,  and  so  on  until  the 
pyramids  which  remain  undivided  in  the  pyramid  DEFH  be, 
all  of  them  together,  less  than  the  excess  of  the  pyramid  DEFH 
above  the  solid  Q:  let  these,  for  example,  be  the  pyramids 
DPRS,  STYH :  therefore  the  prisms,  which  make  the  rest  of 
the  pyramid  DEFH,  are  greater  than  the  solid  Q :  divide  like- 
wise the  pyramid  ABCG  in  the  same  manner,  and  into  as 
many  parts,   as  the   pyramid   DEFH :    therefore,   as  the   base 

b4. 12.  ABC  to  the  base  DEF,  so  •»  are  the  prisms  in  the  pyramid 
ABCG  to  the  prisms  in  the  pyramid  DEFH :  but  as  the  base 
ABC  to  the  base  DEF,  so,  by  hypothesis,  is  the  pyramid  ABCG 
to  the  solid  Q;  and  therefore,  as  the  pyramid  ABCG  to  the 
solid  Q,  so  are  the  prisms  in  the  pyramid  ABC^G  to  the  prisms 
•  in  the   pyramid  DEFH:    but  the   pyramid   ABCG   is   greater 

c  14.  5.  than  the  prisms  contained  in  it;  wherefore  *^  also  the  solid  Q  is 
greater  than  the  prisms  in  the  pyramid  DEFH.  But  it  is  also 
less,  which  is  impossible.     Therefore  the  base  ABC  is  not  to 

•  This  may  be  explained  the  same  way  as  at  the  note  f  in  proposition  2.  in 
the  like  case. 


OF  EUCLID. 


26r 


the  base  DEF,  as  the  pyramid  ABCG  to  any  solid  which  is  less  B.  XII. 
than  the  pyramid  DEFH.  In  the  same  manner  it  may  be  de-  ^■— v-^ 
monstrated,  that  the  base  DEF  is  not  to  the  base  ABC  as  the 
pyramid  DEFH  to  any  solid  which  is  less  than  the  pyramid 
ABCG.  Nor  can  the  base  ABC  be  to  the  base  DEF  as  the  py- 
ramid ABCG  to  any  solid  which  is  greater  than  the  pyra- 
mid DEFH.  For,  if  it  be  possible,  let  it  be  so  to  a  greater,  viz. 
the  solid  Z.  And  because  the  base  ABC  is  to  the  base  DEF  as 
the  pyramid  ABCG  to  the  solid  Z ;  by  inversion,  as  the  base 
DEF  to  the  base  ABC,  so  is  the  solid  Z  to  the  pyramid  ABCG. 
But  as  the  solid  Z  is  to  the  pyramid  ABCG,  so  is  the  pyramid 


B  X  C     E  V  F 


DEFH  to  some  solid*,  which  must  be  less  ^  than  the  pyramid  c  14. 5. 
ABCG,  because  the  solid  Z  is  greater  than  the  pyramid  DEFH. 
And,  therefore,  as  the  base  DEF  to  the  base  ABC,  so  is  the  py- 
ramid DEFH  to  a  solid  less  than  the  pyramid  ABCG ;  the  con- 
trary to  which  has  been  proved.  Therefore  the  base  ABC  is  not 
to  the  base  DEF,  as  the  pyramid  ABCG  to  any  solid  which  is 
greater  than  the  pyramid  DEFH.  And  it  has  been  proved  that 
neither  is  the  base  ABC  to  the  base  DEF  as  the  pyramid  ABCG 
to  any  solid  Avhich  is  less  than  the  pyramid  DEFH.  Therefore, 
as  the  base  ABC  is  to  the  base  DEF,  so  is  the  pyramid  ABCG 
to  the  pyramid  DEFH.     Wherefore,  pyramids,  &c.     Q.  E.  D. 

*  This  may  be  explained  the  same  way  as  the  like  at  the  mark  f  in  prop.  2. 


264 
B.  XII. 


THE  ELEMENTS 


PROP.  VI.    THEOR. 


SeeN.  PYRAMIDS  of  the 
polygons  for  their  bases 
bases. 


same  altitude  which  have 
,  are  to  one  another  as  their 


Let  the  pyramids  which  have  the  polygons  ABCDE,  FGHKL 
for  their  bases,  and  their  vertices  in  the  points  M,  N,  be  of  the 
same  altitude:  as  the  base  ABCDE  to  the  base  FGHKL,  so  is 
the  pyramid  ABCDEM  to  the  pyramid  FGHKLN. 

Divide  the  base  ABCDE  into  the  triangles  ABC,  ACD,  ADE; 
and  the  base  FGHKL  into  the  triangles  FGH,  FHK,  FKL:  and 
upon  the  bases  ABC,  ACD,  ADE  let  there  be  as  many  pyra- 
mids of  which  the  common  vertex  is  the  point  M,  and  upon  the 
remaining  bases  as  many  pyramids  having  their  common  ver- 
tex in  the  point  N:  therefore,  since  the  triangle  ABC  is  to  the 
a  5. 12.  triangle  FGH,  as  ^  the  pyramid  ABCM  to  the  pyramid  FGHN; 
and  the  triangle  ACD  to  the  triangle  FGH,  as  the  pyramid 
ACDM  to  the  pyramid*  FGHN;    and  also  the  triangle  ADE  to 

M  N 


b  2.  Cor. 

24.  5. 


the  triangle  FGH,  as  the  pyramid  .\DEM  to  the  pyramid  FGHN; 
as  all  the  first  antecedents  to  their  common  consequent,  so  ^  are 
all  the  other  antecedents  to  their  common  consequent ;  that  is,  as 
the  base  ABCDE  to  the  base  FGH,  so  is  the  pyramid  ABCDEM 
to  the  pyramid  FGHN :  and,  for  the  same  reason,  as  the  base 
FGHKL  to  the  base  FGH,  so  is  the  pyramid  FGHKLN  to  the 
pyr:imid  FGHN  :  and,  by  inversion,  as  the  base  FGH  to  the  base 
FGHKL,  so  is  the  pyramid  FGHN  to  the  pyramid  FGHKLN:- 
then,  because  as  the  base  ABCDE  to  the  base  FGH,  so  is  the  pyra- 
mid ABCDEM  to  the  pyramid  FGHN ;  and  as  the  base  FGH  to  the 
base  FGHKL,  so  is  the  pyramid  FGHN  to  thi;  pyramid  FGHKLN ; 


OF  EUCLID.  265 

therefore,  ex  gquali  c,  as  the  base  ABCDE  to  the  base  FGHKL,  B.  XII. 
so  the  pyramid  ABCDEM  to  the  pyramid  FGHKLN.  There-  ' — r-^ 
fore,  pyramids,  Sec.     Q.  E.  D.  c  22.  5. 


PROP.  VII.    THEOR.^ 

EVERY  prism  having  a  triangular  base  may  be 
divided  into  three  pyramids  that  have  triangular  bases, 
and  are  equal  to  one  another. 

Let  there  be  a  prism  of  which  the  base  is  the  triangle  ABC, 
and  let  DEF  be  the  triangle  opposite  to  it:  the  prism  ABCDEF 
may  be  divided  into  three  equal  pyramids  having  triangular 
bases. 

Join  BD,  EC,  CD;    and  because   ABED  is  a  parallelogram 
of  which  BD  is  the  diameter,  the  triangle  ABD  is  equal  »  to  a  34. 1. 
the  triangle  EBD  ;    therefore  the  pyramid  of  which  the  base  is 
the  triangle  ABD,  and  vertex  the  point  C,  is  equal  ^  to  the  py-  b  5. 12. 
ramid  of  which  the  base  is  the  triangle  EBD,  and  vertex  the 
point  C  ;    but  this  pyramid  is  the  same  with  the  pyramid  the 
base  of  which  is  the  triangle  EBC,  and  vertex  the  point  D  j    for 
they  are  contained  by  the  same  planes ;    therefore  the  pyramid 
of  Avhich  the  base  is  the  triangle  ABD,  and  vertex  the  point  C, 
is  equal  to  the  pyramid,  the  base  of  which  is  the  triangle  EBC, 
and  vertex  the  point  D :    again,  because  FCBE  is  a  parallelr 
ogram  of  which  the  diameter  is  CE,   the 
triangle   ECF   is   equal  ^  to   the  triangle  F 

ECB  ;  therefore  the  pyramid  of  which 
the  base  is  the  triangle  ECB,  and  vertex 
the  point  D,  is  equal  to  the  pyramid,  the 
base  of  which  is  the  triangle  ECF,  and 
vertex  the  point  D :  but  the  pyramid  of 
which  the  base  is  the  triangle  ECB,  and 
vertex  the  point  D,  has  been  proved  equal 
to  the  pyramid  of  which  the  base  is  the 
triangle    ABD,    and    vertex   the    point   C. 

Therefore  the  prism  ABCDEF  is  divided  into  three  equal  pyra- 
mids having  triangular  bases,  -viz.  inta  the  pyramids  ABDC, 
EBDC,  ECFD  :  and  because  the  pyramid  of  which  tlie  base  is 
the  triangle  ABD,  and  vertex  the  point  C,  is  the  same  with  the 
pyramid  of  which  the  base  is  the  triangle  ABC,  and  vertex  the 
point  D,  for  they  are  contained  by  the  same  planes ;  and  that  the 
pyramid  of  which  the  base  is  the  triangle  ABD,  and  vertex  the 

2  L 


266 


TPTE  EI.EMENTS 


B.  XII.  point  C,  has  been  demonstrated  to  be  a  third  part  of  the  prism 
Wv—V  the  biise  of  which  is  the  triangle  ABC,  and  to  which  DEF  is  the 
opposite  triangle  ;  therefore  the  pyramid  of  which  the  base  is  the 
triangle  ABC,  and  vertex  the  point  D,  is  the  third  part  of  the 
prism  yvhich  has  the  same  base,  viz.  the  triangle  ABC,  and  DEF 
is  the  opposite  triangle.     Q.  E.  D. 

CoR.  1.  From  this  it  is  manifest,  that  every  pyramid  is  the 
third  part  of  a  prism  which  has  tUe  same  base,  and  is  of  an  equal 
altitude  with  it ;  for  if  the  base  of  the  prism  be  any  other  figure 
than  a  triangle,  it  may  be  divided  into  prisms  having  triangular 
bases. 

Cor.  2.    Prisms  of  equal  altitudes  are  to  one  another  as  their 
bases  ;    because  the  pyramids  upon  the  same  bases,  and  of  the 
t  6. 12,    same  altitude,  are  ^  to  one  another  as  their  bases, 


PROP.  VIII.    THEOR, 

SIMILAR  pyramids  having  triangular  bases  are 
one  to  another  in  the  triplicate  ratio  of  that  of  their 
honmologous  sides. 

Let  the  pyramids  having  the  triangles  ABC,  DEF  for  their 
bases,  and  the  points  G,  H  for  their  vertices,  be  similar,  and  si- 
milarly situated;  the  pyramid  ABCG  has  to  the  pyramid  DEFH, 
the  triplicate  ratio  of  that  which  the  side  BC  has  to  the  homolo- 
gous side  EF. 

Complete  the  parallelograms  ABCM,  GBCN,  ABGK,  and 
the  solid  parallelepiped  BGML  contained  by  these  planes  an(i 


B  C  E  F 

those  opposite  to  them  ;  and,  in  like  manner,  complete  the  solid 
parallelepiped  EHPO  contained  by  the  three  parallelograms 
DEFPj  IIEFR,  DEHX,  and  those  opposite  tq  them;   and,  be? 


OF  EUCLID.  2.67 

cause  the  pyramid  ABCG  is  similar  to  the  pyramid  DEFH,  the  B.  Xll. 
angle  ABC  is  equal  ^  to  the  angle  DEF,  and  the  angle  GBC  to  the  v.^-y-^1* 
angle  HEF,  and  ABG  to  DEH :   and  AB  is  b  to  BC,  as  DE  to  EF  ;  a.  11.  def. 
that  is,  the  sides  about  the  equal  angles  are  proportionals  ;  vvhei-e-    !-• 
fore  the  parallelogram  BM  is  similar  to  EP  :  for  the  same  reason,  b  l.def.6. 
the  parallelogram  BN  is  similar  to  ER,  and  BK  to  EX :   there- 
fore the  three  parallelograms  BM,  BN,  BK  are  similar  to  the 
three  EP,  ER,  EX  :  but  the  three  BM,  BN,  BK,  are  equal  and 
similar  =  to  the  three  which  are  opposite  to  them,  and  the  three  c  24  ll. 
EP,  ER,  EX  equal  and  similar  to  the  three  opposite  to  them  : 
wherefore  the  solids  BGML,  EHPO  are  contained  by  the  same 
number  of  similar  planes;  and  their  solid  angles  are  equal  <i ;  d  B.  11. 
and  therefore  the  solid  BGML,  is  similar  ^  to  the  solid  EHPO  : 
but  similar  solid  parallelepipeds  have  the  triplicate  «  ratio  of  that  e  33- 11. 
which  their  homologous  sides  have  :   therefore  the  solid  BGML 
has   to   the  solid  EHPO  the  triplicate  ratio  of  that  which  the 
side  BC  has  to  the  homologous  side  EF  :  but  as  the  solid  BGML 
is  to  the  solid  EHPO,  so  is  f  the  pyramid  ABCG  to  the  pyramid  f  15.  5. 
DEFH  ;  because  the  pyramids  are  the  sixth  part  of  the  solids, 
since  the  prism,  which  is  the  half  s  of  the  solid  parallelepiped,  g  28. 11. 
is  triple  i»  of  the  pyramid.     Wherefoi-e    likewise   the  pyramid  h  7. 12. 
ABCG  has  to  the  pyramid  DEFH,  the  triplicate  ratio  of  that 
which  BC  has  to  the  homologous  side  EF.     Q.  E.  D. 

CoR.  From  this  it  is  evident,  that  similar  pyramids  which  See  Note, 
have  multangular  bases,  are  likewise  to  one  another  in  the  tri- 
plicate ratio  of  their  homologous  sides  :  for  they  may  be  divided 
into  similar  pyramids  having  triangular  bases,  because  the  simi- 
lar polygons,  which  are  their  bases,  may  be  divided  into  the  same 
number  of  similar  triangles  homologous  to  the  whole  polygons  ; 
therefore  as  one  of  the  triangular  pyramids  in  the  first  multan- 
gular pyramid  is  to  one  of  the  triangular  pyramids  in  the  other, 
so  are  all  the  triangular  pyramids  in  the  first  to  all  the  triangu- 
lar pyramids  in  the  other ;  that  is,  so  is  the  first  maltangulaf 
pyramid  to  the  other  :  but  one  triangular  pyramid  is  to  its  simi- 
lar triangular  pyramid,  in  the  triplicate  ratio  of  their  homologous 
sides  ;  and  therefore  the  first  multangular  pyramid  has  to  the 
other  the  triplicate  ratio  of  that  which  one  of  the  sides  of  th« 
first  has  to  the  homologous  side  of  the  other. 


268 
B.  XII. 


THE  ELEMENTS 


PROP.  IX.     THEOR. 


THE  bases  and  altitudes  of  equal  pyramids  having 
triangular  bases  are  reciprocally  proportional  :  and 
triangular  pyramids  of  which  the  bases  and  altitudes 
are  reciprocally  proportional  are  equal  to  one  ano- 
ther. 


I^et  the  pyramids  of  which  the  triangles  ABC,  DEF  are  the 
bases,  and  which  have  their  vertices  in  the  points  G,  H,  be  equal 
to  one  another :  the  bases  and  altitudes  ot*  the  pyramids  ABCG, 
DEFH  are  reciprocally  proportional,  'oiz.  the  base  ABC  is  to  the 
base  DEF,  as  the  altitude  of  the  pyramid  DEFH  ta  the  altitude 
of  the  pyramid  ABCG. 

Complete  the  parallelograms  AC,  AG,  GC,  DF,  DH,  HF ; 
and  the  solid  parallelepipeds  BGML,  EHPO  contained  by  these 


O 


R 


N       y^ 


/ 

H 

/ 

/ 

/ 

P 

\ 

11 

^ 

D  E 


planes  and  those  opposite  to  them  :  and  because  the  pyramid 
ABCG  is  equal  to  the  pyramid  DEFH,  and  that  the  solid  BGML 
is  sextuple  of  the  pyramid  ABCG,  and  the  solid  EHPO  sextuple 

al.Ax.5.  of  the  pyramid  DEFH  ;  therefore  the  solid  BGML  is  equal  *  to 
the  solid  EHPO  :  but  the  bases  and  altitudes  of  equal  solid  pa- 

b34. 11.  rallelepipcds  are  reciprocally  proportional  •*  ;  therefore  as  the 
base  BM  to  the  base  EP,  so  is  the  altitude  of  the  solid  EHPO  to 
the  altitude  of  the  solid  BGML  :  but  as  the  base  BM  to  the  base 

c  15.5.  EP,  so  is  c  the  triangle  ABC  to  the  triangle  DEF  ;  therefore  as 
the  triangle  ABC  to  the  triangle  DEF,  so  is  the  altitude  of  the 
solid  EHPO  to  the  altitude  of  the  solid  BGML :  but  the  altitude 
of  the  solid  EHPO  is  the  same  with  the  altitude  of  the  pyramid 
DEFH  ;  and  the  altitude  of  the  solid  BGML  is  the  same  with  the 


OF  EUCLID.  269 

aUitude  of  the  pyramid  ABCG :  therefore,  as  the  base  ABC  to  B.  XII. 
the  base  DEF,  so  is  the  altitude  of  the  pyramid  DEFH  to  the  ^~— v— ^ 
altitude  of  the  pyramid  ABCG :  wherefore  the  bases  and  alti- 
tudes of  the  pyramids  ABCG,  DEFH  are  reciprocally  propor- 
tional. 

Again,  Let  the  bases  and  altitudes  of  the  pyramids  ABCG, 
DEFH  be  reciprocally  proportional,  viz.  the  base  ABC  to  the 
base  DEF,  as  the  altitude  of  the  pyramid  DEFH  to  the  altiude 
of  the  pyramid  ABCG :  the  pyramid  ABCG  is  equal  to  the  py- 
ramid DEFPL 

The  same  construction  being  made,  because  as  the  base  AB(' 
to  the  base  DEF,  so  is  the  altitude  of  the  pyramid  DEFH  to  the 
altitude  of  the  pyramid  ABCG  :  and  as  the  base  ABC  to  the  base 
DEF,  so  is  the  parallelogram  BM  to  the  parallelogram  EP : 
therefore  the  parallelograni  BM  is  to  EP,  as  the  altitude  of  the 
pyramid  DEFH  to  the  altitude  of  the  pyramid  ABCG :  but  the 
altitude  of  the  pyramid  DEFH  is  the  same  with  the  altitude  of 
the  solid  parallelepiped  EHPO  ;  and  the  altitude  of  the  pyramid 
ABCG  is  the  same  with  the  altitude  of  the  solid  parallelepiped 
BGML :  as,  therefore,  the  base  BM  to  the  base  EP,  so  is  the 
altitude  of  the  solid  parallelepiped  EHPO  to  the  altitude  of  the 
solid  parallelepiped  BGML.  But  solid  parallelepipeds,  having 
their  bases  and  altitudes  reciprocally  proportional,  are  equal  *>  to  b  34.  II. 
one  another.  Therefore  the  solid  parallelepiped  BGML  is  equal 
to  the  solid  parallelepiped  EHPO.  And  the  pyramid  ABCG  is 
the  sixth  part  of  the  solid  BGML,  and  the  pyramid  DEFH  is  the 
sixth  part  of  the  solid  EHPO.  Therefore  the  pyramid  ABCG  is 
equal  to  the  pyramid  DEFH.   Therefore  the  bases,  £cc.  Q.  E.  D. 


PROP.  X.    THEOR. 

EVERY  cone  is  the  third  part  of  a  cylinder  which 
has  the  same  base,  and  is  of  an  equal  altitude  with  it. 

Let  a  cone  have  the  same  base  with  a  cylinder,  viz.  the  circle 
ABCD,  and  the  same  altitude.  The  cone  is  the  third  part  of  the 
cylinder ;  that  is,  the  cylinder  is  triple  of  the  cone. 

If  the  cylinder  be  not  triple  of  the  cone,  it  mnst  either  be 
greater  than  the  triple,  or  less  than  it.  First,  Let  it  be  greater 
than  the  triple ;  and  describe  the  square  ABCD  in  the  circle  ; 
this   square   is   greater  than   the   half  of  the   circle   ABCD*. 

*  As  was  shown  in  prop.  2-  of  this  book. 


sro  THE  ELEMENTS 

B.  XIl.  Upon  the  square  ABCD  erect  a  prism  of  the  same  dltitude  with 

^•^y-*"^  the  cylinder;  this  prism  is  p^reater  than  half  of  the  cylinder; 
because  if  a  square  be  described  about  the  circle,  and  a  prism 
erected  upon  the  square,  of  the  same  altitude  with  the  cylinder, 
the  inscribed  square  is  half  of  that  circumscribed;  and  upon 
these  square  bases  are  erected  solid  parallelepipeds,  viz,  the 
prisms  of  the  same  altitude ;  therefore  the  prism  upon  the 
square  ABCD  is  the  half  of  the  prism  upon  the  square  descri- 
bed about  the  circle :  because  they  are  to  one  another  as  their 

a  32.  11.  bases  ^ .-  and  the  cylinder  is  less  than  the  prism  upon  the  square 
described  about  the  circle  ABCD :  therefore  the  prism  upon 
the  square  ABCD  of  the  same  altitude  with  the  cylinder,  is 
greater  than  half  of  the  cylinder.  Bisect  the  circumferences 
AB,  BC,  CD,  DA,  in  the  points  E,  F,  G,  H ;  and  join  AE,  EB, 
BF,  FC,  CG,  GD,  DH,  HA :  then,  each  of  the  triangles  AEB, 
BFC,  CGD,  DHA  is  greater  than  the  half  of  the  segment  of 
the  circle  in  which  it  stands,  as  was  ^ 

shown  in  prop  2.  of  this  book. 
Erect  prisms  upon  each  of  these 
triangles  of  the  same  altitude  with 
the  cylinder ;  each  of  these  prisms 
is  greater  than  half  of  the  segment 
of  the  cylinder  in  which  it  is :  be- 
cause if,  through  the  points  E,  F,  G, 
H,  parallels  be  drawn  to  AB,  BC, 
CD,  DA,  and  parallelograms  be 
completed  upon  the  same  AB,  BC, 
CD,  DA,  and  solid  parallelepipeds 
be  erected  upon  the  parallelograms ;  the  prisms  upon  the 
triangles  AEB,  BFC,  CGD,  DIL'V  arc  the  halves  of  the  solid 

b  2  Cor.  parallelepipeds  ^.  And  the  segments  of  the  cylinder  Avhich  are 
7- 12.  upon  the  segments  of  the  circle  cut  off  by  AB,  BC,  CD,  DA, 
are  less  than  the  solid  parallek])ipeds  which  contain  them. 
Therefore  the  prisms  upon  the  triangles  AEB,  BFC,  CGD, 
DHA,  are  greater  than  half  of  the  segments  of  the  cylinder  in 
which  they  are;  therefore,  if  each  of  the  circumferences  be  di- 
vided into  two  equal  parts,  and  straight  lines  be  drawn  from 
the  points  of  division  to  the  extremities  of  the  circumferences, 
and  upon  the  triangles  thus  made,  prisms  be  erected  of  the  same 

c  Lem-    altitude  with  the  cylinder,  and  so  on,  there  must  at  length  re- 

mu..         main  some  segments  of  the  cylinder  which  together  are  less  <=■ 

than  the  excess  of  the  cylinder   above  the  triple   of  the   cone. 

Let  ihem  be  those  upon  the  segments  of  tlic  circle  AE,  EB,  ]J1', 


OF  EUCLID. 


271 


FC,  CG,  GD,  DH,  HA.     Therefore  the  rest  of  the  cylinder,  that  B.  XII. 
is,  the  prism  of  which  the  base  is  the  polys^on  AEBFCGDH,  and  ^  —^.mJ 
of  which  the  altitude  is  the  same  with  that  of  the  cylinder,  is 
greater  than  the  triple  of  the  cone  :    but  this  prism  is  triple  ^  of  d  1.  Cor. 
the  pyramid  upon  the  same  base,  of  which  the  vertex  is  the  same  7-  12- 
with  the  vertex  of  the  cone  ;  therefore  the  pyramid  upon  the  base 
AEBFCGDH,  having  the  same  vertex  with  the  cone,  is  greater 
than  the  cone,  of  which  the  base  is  the  circle  ABCD :    but  it  is 
also  less,  for  the  pyramid  is  contained  within  the  cone  ;  which  is 
impossible.     Nor  can  the  cylinder  be  less  than  the  triple  of  the 
cone.     Let  it  be  less,  if  possible  :    therefore,  inversely,  the  cone 
is  greater  than  the  third  part  of  the  cylinder.    In  the  circle  ABCD 
describe  a  square ;    this  square  is  greater  than  the  half  of  the 
circle :   and  upon  the  square  ABCD  erect  a  pyramid  having  the 
same  vertex  with  the  cone  ;    this  pyramid  is  greater  than  the 
half  of  the  cone  ;  because,  as  was  before  demonstrated,  if  a  square 
be  described   about  the   circle,  the 
square  ABCD  is  the  half  of  it;  and 
if,  upon  these  squares,  there  be  erect- 
ed solid  parallelepipeds  of  the  same 
altitude  with  the  cone,  which  are  also 
prisms,  the  prism  upon  the  square 
ABCD  shall  be  the  half  of  that  which 
is  upon  the  square  described  about 
the  circle  ;  ior  they  are  to  one  ano- 
ther as  their  bases*  ;  as  are  also  the  H- ! /  a  32. 11. 

third  parts  of  them  :  therefore  the 
pyramid,  the  base  of  which  is  the 
square  ABCD,  is  half  of  the  pyramid 

upon  the  square  described  about  the' circle  :  but  this  last  pyramid 
is  greater  than  the  cone  which  it  contains  ;  therefore  the  pyramid 
upon  the  square  ABCD,  having  the  same  vertex  with  the  cone, 
is  greater  than  the  half  of  the  cone.  Bisect  the  circumferences 
AB,  BC,  CD,  DA  in  the  points  E,  F,  G,  H,  and  join  AE,  EB,  BF, 
FC,  CG,  GD,  DH,  HA:  therefore  each  of  the  triangles  AEB, 
BFC,  CGD,  DHA  is  greater  than  half  of  the  segment  of  the  cir- 
cle in  which  it  is:  upon  each  of  these  triangles  erect  pyramids 
having  the  same  vertex  with  the  cone.  Therefore  each  of  these 
pyramids  is  greater  than  the  half  of  the  segment  of  the  cone  in 
Avhich  it  is,  as  before  was  demonstrated  of  the  prisms  and  seg- 
ments of  the  cylinder  ;  and  thus  dividing  each  of  the  circumfe- 
rences into  two  equal  parts,  and  joining  the  points  of  division  and 
their  extremities  by  straight  lines,  and  upon  the  triangles  erect- 
ing pyramids  having  their  vertices  the  same  with  that  of  the  cone, 
?ind  so  on,  there  must  at  length  remain  some  segments  of  the 
fone,  which  together  shall  be  less  than  the  excess  of  the  cone 


272 


THE  ELEMENTS 


H 


B.  XII.  above  the  third  part  of  the  cyHnder.  Let  these  be  the  segments 
V— V— '  upon  AE,  EB,  BF,  FC,  CG,  GD,  DH,  HA.  Therefore  the  rest 
of  the  cone,  that  is,  the  pyramid, 
of  which  the  base  is  the  polygon 
AEBFCGDH,  and  of  which  the  ver- 
tex is  the  same  with  that  of  the  cone, 
is  greater  than  the  third  part  of  the 
cylinder.  But  this  pyramid  is  the 
third  part  of  the  prism  upon  the  same 
base  AEBFCGDH,  and  of  the  same 
altitude  with  the  cylinder.  Therefore 
this  prism  is  greater  than  the  cylinder 
of  which  the  base  is  the  circle  ABCD. 
But  it  is  also  less,  for  it  is  contained 
within  the  cylinder  ;  which  is  impos- 
sible. Therefore  the  cylinder  is  not  less  than  the  triple  of  the 
cone.  And  it  has  been  demonstrated  that  neither  is  it  greater 
than  the  triple.  Therefore  the  cylinder  is  triple  of  the  cone,  or, 
the  cone  is  the  third  part  of  the  cylinder.  Wherefore,  eveiy 
'  one,  Sec.     Q.  E.  D. 


PROP.  XL    THEOR. 


See  N.        CONES  and  cylinders  of  the  same  altitude  are  to 
one  another  as  their  bases. 


Let  the  cones  and  cylinders,  of  which  the  bases  are  the  circles 
ABCD,  EFGH,  and  the  axes  KL,  MN,  and  AC,  EG  the  diame- 
ters of  their  bases,  be  of  the  same  altitude.  As  the  circle  ABCD 
to  the  circle  EFGH,  so  is  the  cone  AL  to  the  cone  EN. 

If  it  be  not  so,  let  the  circle  ABCD  be  to  the  circle  EFGH, 
as  the  cone  AL  to  some  solid  either  less  than  the  cone  EN,  or 
greater  than  it.  First,  let  it  be  to  a  solid  less  than  EN,  viz.  to 
the  solid  X ;  and  let  Z  be  the  solid  which  is  equal  to  the  ex- 
cess of  the  cone  EN  above  the  solid  X;  therefore  the  cone  EN 
is  equal  to  the  solids  X,  Z  together.  In  the  circle  EFGH  de- 
scribe tlie  square  EFGFI,  therefore  this  square  is  greater  than 
tlie  half  of  the  circle :  upon  the  square  EFGH  erect  a  pyra- 
mid of  ihe  same  altitude  with  tlie  cone  ;  this  pyramid  is  greater 
than  half  of  the  cone.  For,  if  a  square  be  described  about  th^ 
circle,  and  a  pyramid  be  erected  upon  it,  having  the  same  ver- 


OF  EUCLID. 


27a. 


tex  with  the  cone*,  the  pyramid  inscribed  in  the  cone  is  half  B.  XII. 
of  the  pyramid  circums,cribed  about  it,  because  they  are  to  one  ^  —^mj 
another  as  their  bases  » :   but  the  cone  is  less  than  the  circum-  a  6. 12. 
scribed  pyramid  ;  therefore  the  pyramid  of  which  the  base  is  the 
square  EFGH,  and  its  vertex  the  same  with  that  of  the  cone,  is 
greater  than  half  of  the  cone :    divide  the  circumferences  EF, 
FG,  GH,  HE,  each  into  two  equal  parts  in  the  points  O,  P,  R, 
S,  and  join  EO,  OF,  FP,  PG,  GR,  RH,  HS,  SE :  therefore  each 
of  the  triangles  EOF,  FPG,  GRH,  HSE  is  greater  than  half  of 


the  segment  of  the  circle  in  which  it  is:  upon  each  of  these 
triangles  erect  a  pyramid  having  the  same  vertex  with  the  cone  ; 
each  of  these  pyramids  is  greater  than  the  half  of  the  segment 
of  the  cone  in  which  it  is :  and  thus,  dividing  each  of  these  cir- 
cumferences into  two  equal  parts,  and  from  the  points  of  division 
drawing  straight  lines  to  the  extremities  of  the  circumferences, 
and  upon  each  of  the  triangles  thus  made  erecting  pyran)ids 
having  the  same  vertex  with  the  cone,  and  so  on,  there  must  at 
length  remain  some  segments  of  the  cone  which  are  togt;ther 
less''  than  the  solid  Z  :  let  these  be  the  segments  upon  EO,  OF, bLcm.l. 

*  Vertex  is  put  in  place  of  ahitude  which  is  in  the  Greek,  because  the  pyra- 
mid, in  what  follows,  is  supposed  to  be  circumscribed  about  the  cone,  and  so 
must  have  the  same  vertex.  And  the  same  change  is  made  m  some  places  fol- 
lowing. 

3  M 


274 


THE  ELEMENTS 


B.  XII.  FP,  PG,  GR,  RH,  HS,  SE  :  therefore  the  remainder  of  the  cone,^ 

^■*-»-^-^  viz.  the  pyramid  of  which  the  base  is  the  polygon  EOFPGRHS, 

and  its  vertex  the  same  with  that  of  the  cone,  is  greater  than 

the    solid    X:    in    the    circle    ABCD    describe    the    polygon 

ATBYCVDQ  similar  to  the  polygon  EOFPGRHS,  and  upon 

it  erect  a  pyramid  having  the  same  vertex  with  the  cone  AL : 

a  1. 12.     and  because  as  the  square  of  AC  is  to  the  square  of  EG,  so  ^  is 

the  polygon  ATBYCVDQ  to  the  polygon  EOFPGRHS;  and 

b  2. 12.     as  the   square  of  AC  to  tlie  square  of  EG,   so  is  ^  the  circle 

c  11.  $.     ABCD,  to  the  circle  EFGH  ;  therefore  the  circle  ABCD  is  <=  to 

the  circle  EFGH,  as  the  polygon  ATBYCVDQ  to  the  poly- 


\ 

N 

I 

\ 

\l 

gon  EOFPGRHS :  but  as  the  circle  ABCD  to  the  circle  EFGH, 
so  is  the  cone   AL  to  the  the   selid  X  ;   and   as  the  polygon 

d  6.  12.  ATBYCVDQ  to  the  polygon  EOFPGRHS,  so  is  <^  the  pyra- 
mid of  which  the  base  is  the  first  of  these  polygons,  and  ver- 
tex L,  to  the  pyramid  of  which  the  base  is  the  other  polygon, 
and  its  vertex  N :  therefore,  as  the  cone  AL  to  the  solid  X,  so 
is  the  pyramid  of  which  the  base  is  the  polygon  ATBYCVDQ, 
and  vertex  L,  to  the  pyramid  the  base  of  which  is  the  polygon 
EOFPGRHS,  and  vertex  N:  but  the  cone  AL  is  greater  than 

eU.  5.  the  pyramid  contained  in  it;  therefore  the  solid  X  is  greater* 
than  the  pyramid  in  the  cone  EN  ;  but  it  is  less,  as  was  shown, 


OF  EUCLID.  S75 

Vhich  is  absurd :  therefore  the  circle  ABCD  is  not  to  the  circle  B.  XII. 
EFGH  as  the  cone  AL  to  any  solid  which  is  less  than  the  cone  ^^^-^m^ 
EN.    In  the  same  manner  it  may  be  demonstrated  that  the  circle 
EFGH  is  not  to  the  circle  ABCD  as  the  cone  EN  to  any  solid 
less  than  the  cone  AL.     Nor  can  the  circle  ABCD  be  to  the  cir- 
cle EFGH  as  the  cone  AL  to  any  solid  greater  than  the  cone 
EN:    for,  if  it  be  possible,  let  it  be  so  to  the  solid  I,  which  is 
greater  than  the  cone  EN :  therefore,  by  inversion,  as  the  circle 
EFGH  to  the  circle  ABCD,  so  is  the  solid  I  to  the  cone  AL  :  but 
as  the  solid  I  to  the  cone  AL,  so  is  the  cone  EN  to  some  solid, 
which  must  be  less  *  than  the  cone  AL,  because  the  solid  I  is  a  14.  5- 
greater  than  the  cone  EN :    therefore,  as  the  circle  EFGH  is  to 
the  circle  ABCD,  so  is  the  cone  EN  to  a  solid  less  than  the  cone 
AL,   which  was  shown  to  be  impossible :    therefore  the  circle 
ABCD  IS  not  to  the  circle  EFGH  as  the  cone  AL  is  to  any  solid 
greater  than  the  cone  EN :    and  it  has  been  demonstrated  that 
neither  is  the  circle  ABCD  to  the  circle  EFGH,  as  the  cone  AL 
to  any  solid  less  than  the  cone  EN  :  therefore  the  circle  ABCD  is 
to  the  circle  EFGH,  as  the  cone  AL  to  the  cone  EN:  but  as  the 
cone  is  to  the  cone,  so  ^  is  the  cylinder  to  the  cylinder,  because  b  15.  5. 
the  cylinders  are  triple  <=  of  the  cone,  each  to  each.     Therefore,  c  10. 1% 
as  the  circle  ABCD  to  the  circle  EFGH,  so  are  the  cylinders  up- 
on them  of  the  same  altitude.     Wherefore,  cones  and  cylinders 
of  the  same  altitude  are  to  one  another  as  their  bases.    Q.  E.  D. 


PROP.  Xn.    THEOR. 


SIMILAR  cones  and  cylinders  have  to  one  ano-  see  n. 
ther  the  triplicate  ratio  of  that  which  the  diameters  of 
their  bases  have. 


Let  the  cones  and  cylinders  of  which  the  bases  are  the  circles 
ABCD,  EFGH,  and  the  diameters  of  the  bases  AC,  EG,  and 
KL,  MN  the  axis  of  the  cones  or  cylinders,  be  similar :  the  cone 
of  which  the  base  is  the  circle  ABCD,  and  vertex  the  point  L, 
has  to  the  cone  of  which  the  base  is  the  circle  EFGH,  and  ver- 
tex N,  the  triplicate  ratio  of  that  which  AC  has  to  EG. 

For,  if  the  cone  ABCDL  has  not  to  the  cone  EFGHN  the  tri- 
plicate ratio  of  that  which  AC  has  to  EG,  the  cone  ABCDL  shall 
have  the  triplicate  of  that  ratio  to  some  «olid  which  is  less  or 


276 


THE  ELEMENTS 


B.  XII.  greater  than  the  cone  EFGHN.  First,  let  it  have  it  to  a  less, 
^>— v-^  viz.  to  the  solid  X.  Make  the  same  construction  as  in  the  pre- 
ceding proposition,  and  it  may  be  demonstrated  the  very  same 
way  as  in  that  proposition,  that  the  pyramid  of  which  the  base 
is  the  polyi^on  EOFPGRHS,  and  vertex  N,  is  greater  than  the 
soHd  X.  Describe  also  in  the  circle  ABCD  the  polygon 
ATBYCv'DQ  simi  ar  to  the  polygon  EOFPGRHS,  upon  which 
erect  a  pyramid  having  the  same  vertex  with  the  cone ;  and  let 
LAQ  be  one  of  the  triangles  containing  the  pyramid  upon 
the  polygon  ATBYCVDQ,  the  vertex  of  which  is  L ;  and  let 
NES  be  one  of  the  triangles  containing  the  pyramid  upon  the 


polygon  EOFPGRHS,  of  which  the  vertex  is  N ;  and  join  KQ, 
MS:    because  then  the  cone  ABCDL  is  similar  to  the  cone 

a  24.  def.  EFGHN,  AC  is  »  to  EG,  as  the  axis  KL  to  the  axis  MN ;  and 
11.        as  AC  to  EG,  so  b  is  AK  to  EM  ;    therefore  as  AK  to  EM, 

b  15.  5.  so  is  KL  to  MN ;  and,  alternately,  AK  to  KL,  as  EM  to 
MN:  and  the  right  angles  AKL,  EMN  are  equal;  therefore 
the    sides    about   these    equal    angles   being   proportionals,   the 

c  6.  6.  triangle  AKL  is  similar  <=  to  the  triangle  EMN.  Again,  be- 
cause AK  is  to  KQ,  as  EM  to  MS,  and  that  these  sides  are 


OF  EUCLID.  2rr 

about  equal  angles  AKQ,  EMS,  because  these  angles  are,  each  B.  XII, 
of  them,  the  same  part  of  four  right  angles  at  the  centres  K,  M  ;  ^  -y— ^ 
therefore  the  triangle  AKQ  is  similar  *  to  the  triangle  EMS  :  a  6.  §. 
and  because  it  has  been  shown  that  as  AK  to  KL,  so  is  EM  to 
MN,  and  that  AK  is  equal  to  KQ,  and  EM  to  MS,  as  QK  to 
KL,  so  is  SM  to  MN ;  and  therefore  the  sides  about  the  right 
angles  QKL,  SMN  being  proportionals,  the  triangle  LKQ  is  si- 
milar to  the  triangle  NMS  :  and  because  of  the  similarity  of  the 
triangles  AKL,  EMN,  as  LA  is  to  AK,  so  is  NE  to  EM;   and 
by  the  similarity  of  the  triangles  AKQ,  EMS,  as  KA  to  AQ,  so 
ME  to  ES  ;  ex  aquali^,  LA  is  to  AQ,  as  NE  to  ES.     Again,  be-  b  22-  5. 
cause  of  the  similarity  of  the  triangles  LQK,  NSM,  as  LQ  to 
QK,  so  NS  to  SM :   and   from  the  similarity  of  the  triangles 
KAQ,  MES,  as  KQ  to  QA,  so  MS  to  SE ;  ex  xqvali  b,  LQ  is  to 
QA,  as  NS  to  SE :  and  it  was  proved  that  QA  is  to  AL,  as  SE 
to  EN  ;  therefore,  again,  ex  aquali,  as  QL  to  LA,  so  is  SN  to 
NE  :  wherefore  the  triangles  LQA,  NSE,  having  the  sides  about 
all  their  angles  proportionals,  are  equiangular  =  and  similar  to  c  5.  6. 
one  another  :   and  therefore  the  pyramid  of  which  the  base  is 
the  triangle  AKQ,  and  vertex  L,  is  similar  to  the  pyramid  the 
base  of  which  is  the  triangle  EMS,  and  vertex  N,  because  their 
solid  angles  are  equal  ^  to  one  another,  and  they  are  contained  d  B.  11. 
by  the  same  number  of  similar  planes  :   but  similar  pyramids 
which  have  triangular  bases  have  to  one  another  the  triplicate 
«  ratio  of  that   which   their  homologous  sides  have ;   therefore  e  8. 12. 
the  pyramid  AKQL  has  to  the  pyramid  EMSN  the  triplicate 
ratio  of  that  which  AK  has  to  EM.     In  the  same  manner,  if 
straight  lines    be    drawn   from  the  points   D,  V,  C,  Y,  B,  T 
to  K,  and  from  the  points  H,  R,  G,  P,  F,  O  to  M,  and  py- 
ramids be  erected  upon  the  triangles  having  the  same  vertices 
v/ith  the  cones,  it  may  be  demonstrated  that  each  pyramid  in 
the  first  cone  has  to  each  in  the  other,  taking  them  in  the  same 
order,  the  triplicate  ratio  of  that  Avhich  the  side  AK  has  to  the 
side  EM;   that  is,  which  AC  has  to  EG:   but  as  one   antece- 
dent to  its  consequent,  so  are   all   the   antecedents  to  all  the 
consequents  f ;  therefore   as  the  pyramid  AKQL  to   the  pyra-fl2-5. 
mid  EMSN,  so  is  the  whole  pyramid  the  base  of  which  is  the 
polygon    DQATBYCV,  and  vertex   L,  to  the   whole  pyramid 
of  which  the  base  is  the  polygon  HSEOFPGR,  and  vertex  N. 
Wherefore  also  the  first  of  these  two  last  named  pyramids  has 
to  the  other  the  triplicate  ratio  of  that  which  AC  has  to  EG. 
But,  by  the  hypothesis,  the  cone  of  which  the  base  is  the  cir- 
cle   ABCD,    and   vertex  L,   has   to  the   solid  X,  the  triplicate  j 
ratio  of  that  which  AC  has  to  EG ;  therefore,  as  the  cone  of  | 
which  the  base  is  the  circle  ABCD   and  vertex   L,   is  to  the 


278 


THE  ELEMENT^ 


B.  XII.  solid  X,  s'o  is  tire  pyramid  the  base  of  which  Is  the  polygon 
^'-y-^  DQATBYCV,  and  vertex  L,  to  the  pyramid  the  base  of  which 
is  the  polygon  HSEOFPGR  and  vertex  N:  but  the  said  cone 
is  greater  than  the  pyramid  contained  in  it,  therefore  the  solid 
a  H.  5.  X  is  greater  ^  than  the  pyramid,  the  base  of  which  is  the  poly- 
gon HSEOFPGR,  and  vertex  N  ;  but  it  is  also  less  ;  which  is 
impossible :   therefore  the  cone  of  which  the  base  is  the  circle 


ABCD,  and  vertex  L,  has  not  to  any  solid  which  is  less  than  the 
cone  of  which  the  base  is  the  circle  EFGH  and  vertex  N,  the  tri- 
plicate ratio  of  that  which  AC  has  to  EG.  In  the  same  manner 
it  may  be  demonstrated  that  neither  has  the  cone  EFGHN  to 
any  solid  which  is  less  than  the  cone  ABCDL,  the  triplicate 
ratio  of  that  which  E(i  has  to  AC.  Nor  can  the  cone  ABCDL 
have  to  any  solid  which  is  greater  than  the  cone  EFGHN,  the 
triplicate  ratio  of  that  which  AC  has  to  EG  :  for,  if  it  be  possi- 
ble, let  it  have  it  to  a  greater,  viz.  to  the  solid  Z  :  therefore, 
inversely,  the  solid  Z  has  to  the  cone  ABCDL,  the  triplicate 
ratio  of  that  which  EG  has  to  AC  :  but  as  the  solid  Z  is  to 


OF  EUCLID. 


279 


the  cone   ABCDL,   so  is   the   cone   EFGHN  to  some   solid,  B.  XII. 
which  must  be  less  ^  than  the  cone  ABCDL,  because  the  solid  *— v-— ' 
Z  is  greater  than  the  cone  EFGHN  :  therefore  the  cone  EFGHN  a  14.  5. 
has  to  a  solid  which  is  less  than  the  cone  ABCDL,  the  tripli- 
cate ratio  of  that  which  EG  has  to  AC,  which  was  demonstrat- 
ed to  be  impossible :    therefore  the  cone  ABCDL  has  not  to  any 
solid  greater  than  the  cone  EFGHN  the  triplicate  ratio  of  that 
which  AC  has  to  EG :  and  it  was  demonstrated  that  it  could  not 
have  that  ratio  to  any  solid  less  than  the  cone  EFGHN:    there- 
fore the  cone  ABCDL  has  to  the  cone  EFGHN  the  triplicate 
ratio  of  that  which  AC  has  to  EG :   but  as  the  cone  is  to  the 
cone,  so  ^  is  the  cylinder  to  the  cylinder ;   for  every  cone  is  the  b  15.  5 
third  part  of  the  cylinder  upon  the  same  base,  and  of  the  same 
altitude :  therefore  also  the  cylinder  has  to  the  cylinder  the  tri- 
plicate ratio  of  that  which  AC  has  to  EG.     Wherefore,  similar 
cones,  £cc.     Q.  E.  D. 


PROP.  Xni,    THEOR. 

IF  a  cylinder  be  cut  by  a  plane  parallel  to  its  op-  See  n 
posite  planes,  or  bases ;    it  divides  the  cylinder  into 
two  cylinders,  one  of  which  is  to  the  other  as  the 
axis  of  the  first  to  the  axis  of  the  other. 


Let  the  cylinder  AD  be  cut  by  the 
plane  GH,  parallel  to  the  opposite 
planes  AB,  CD,  meeting  the  axis 
EF  in  the  point  K,  and  let  the  line 
GH  be  the  common  section  of  the 
plane  GH  and  the  surface  of  the  cy- 
linder AD :  let  AEFC  be  the  paral- 
lelogram, in  any  position  of  it,  by  the 
revolution  of  which  about  the  straieht 
line  EF  the  cyHnder  AD  is  described; 
and  let  GK  be  the  common  section 
of  the  plane  GH,  and  the  plane 
AEFC  :  and  because  the  parallel 
planes  AB,  GH  are  cut  by  the  plane 
AEKG,  AE,  KG,  their  common  sec- 
tions with  it,  are  parallel »  ;  where- 
fore AK  is  a  parallelogram,  and  GK 
equal  to  EA  the  straight  line  from 
the  centre  of  the  circle  AB :  for  the 
s^rae  reason,  eaph  of  ^he  straight  lines 


R 


B 

H 

D      a  16.  l:i. 

Y 

Q 


280 


THE  ELEMENTS 


R 


B.  XII.  drawn  from  the  point  K  to  the  line  GH  may  be  proved  to  be 

*— V— ^  equal  to  those  which  are  drawn  from  the  centre  of  the  circle 
AB  to  its  circumference,  and  are  therefr-re  all  equal  to  one  uno- 

al5.  def,  ther.  Therefore  the  line  GH  is  the  circumference  of  a  circle  *, 
^'  of  which  the  centre  is  the  point  K:  therefore  the  piane  GH  di- 
vides the  cylinder  AD  into  the  cylinders  AH,  GD ;  for  they  are 
the  same  which  would  be  described  by  the  revolution  of  the 
parallelograms  AK,  GF,  about  the  straight  lines  EK,  KF  :  and 
it  is  to  be  shown,  that  the  cylinder  AH  is  to  the  cylinder  HC,  as 
the  axis  EK  to  the  axis  KF. 

Produce  the  axis  EF  both  ways ;  and  take  any  number  oC 
straight  lines  EN,  NL,  each  equal  to  EK ;  and  any  number  r  X, 
XM,  each  equal  to  FK  ;  and  let  planes 
parallel  to  AB,CD  pass  through  the  points 
L,  N,  X,  M  :  therefore  the  common  sec- 
tions of  these  planes  with  the  cylinder 
produced  are  circles,  the  centres  of  which 
are  the  points  L,  N,  X,  M,  as  was  proved 
of  the  plane  GH ;  and  these  planes  cut 
off  the  cylinders  PR,  RB,  DT,  TQ  :  and 
because  the  axes  LN,  NE,  EK  are  all 
equal,  therefore  the   cylinders  PR,  RB, 

h  11. 12.  BG  are  ^  to  one  another  as  their  bases  ; 
but  their  bases  are  equal,  and  therefore 
the  cylinders  PR,  RB,  BG  are  equal :  and 
because  the  axes  LN,  NE,  EK  are  equal 
to  one  another,  as  also  the  cylinders  PR, 
RB,  BG,  and  that  there  are  as  many  axes 
as  cylinders  ;  therefore,  whatever  multi- 
ple the  axis  KL  is  of  the  axis  KE,  the 
same  multiple  is  the  cylinder  PG  of 
the  cylinder  GB :  for  the  same  reason, 
whatever  multiple  the  axis  MK  is  of  the 
axis  KF,  the  same  multiple  is  the  cylin- 
der QG  of  the  cylinder  GD  :  and  if  the  axis  KL  be  equal  to  the 
axis  KM,  the  cylinder  PG  is  equal  to  the  cylinder  GQ ;  and  if 
the  axis  KL  be  greater  than  the  axis  KM,  the  cylinder  PG  is 
greater  than  the  cylinder  QG ;  and  if  less,  less:  since  there- 
fore there  are  four  magnitudes,  viz.  the  axes  EK,  KF,  and 
the  cylinders  BG,  GD,  and  that  of  tlie  axis  EK  and  cylindei' 
yG,  there  has  been  taken  any  equimultiples  whatever,  viz,  th^ 


G 


V 


B 


H 


D 


Q 


OF  EUCLID. 


281 


axis  KL  and  cylinder  PG ;  and  of  the  axis  KF  and  cylinder  GD,  B.  XII. 
any  equimultiples  whatever,  viz.  the  axis  KM  and  cylinder  GQ :  *-— v*^ 
and  it  has  been  demonsti'ated,  if  the  axis  KL  be  greater  than  the 
axis  KM,  the  cylinder  PG  is  greater  than  the  cylinder  GQ ;  and 
if  equal,  equal ;   and  if  less,  less  :   therefore  ^  the  axis  EK  is  to  d  5.  def. 
the  axis  KF,  as  the  cylinder  BG  to  the  cylinder  GD.     Where-   ^■ 
fore,  if  a  cylinder,  8cc.     Q.  E.  D. 


PROP.  XIV.    THEOR. 

CONES  and  cylinders  upon  equal  bases  are  to 
one  another  as  their  altitudes. 


Let  the  cylinders  EB,  FD  be  upon  the  equal  bases  AB,  CD : 
as  the  cylinder  EB  to  the  cylinder  FD,  so  is  the  axis  GH  to  the 
axis  KL. 

Produce  the  axis,  KL  to  the  point  N,  and  make  LN  equal 
to  the  axis  GH,  and  let  CM  be  a  cylinder  of  which  the  base  is 
CD,  and  axis  LN,  and  because  the   cylinders   EB,  CM   have 
the  same  altitude,  they  are  to  one  another  as  their  bases  3^:    but  a  IJ.  12. 
their  bases  are  equal,  therefore  also  the  cylinders  EB,  CM  are 
equal.     And  because  the  cylin- 
der FM   is  cut  by  the   plane  F/     ^ 
CD    parallel    to    its    opposite 
planes,  as  the  cylinder  CM  to 
the  cylinder  FD,    so  is  **  the 
axis  LN  to  the  axis  KL.     But 
the  cylinder  CM   is  equal  to 
the  cylinder  EB,  and  the  axis 
LN  to  the  axis  GH  :  therefore 
as  the  cylinder  EB  to  the  cylin- 
der FD,  so  is  the  axis  GH  to 
the  axis  KL  :  and  as  the  cylin- 
der EB  to  the  cylinder  FD,  so 
is  c  the  cone  ABG  to  the  cone  CDK,  because  the  cylinders  are  c  15.  5. 
triple  ^  of  the  cones :    therefore  also  the  axis  GH  is  to  the  axis  d  10. 12 
KL,  as  the  cone  ABG  to  the  cone  CDK,  and  the  cylinder  EB  to 
the  cylinder  FD.     Wherefore,  cones,  &c.    Q.  E.  D. 


b  13- 12, 


2N 


282 
B.  XII. 


THE  ELEMENTS 


PROP.  XV.  THEOR. 


See  N.        THE  bases  and  altitudes  of  equal  cones  and  cy- 
linders are  reciprocally  proportional,  and,  if  the  bases 
and  altitudes  be  reciprocally  proportional,  the  cones^ 
and  cylinders  are  equal  to  one  another. 

Let  the  circles  ABCD,  EFGH,  the  diameters  of  which  are 
Ac,  EG,  be  the  bases,  and  KL,  MN  the  axis,  as  also  the  alti- 
tudes, of  equal  cones  and  cylinders ;  and  let  ALC,  ENG  be  the 
cones,  and  AX,  EO  the  cylinders :  the  bases  and  altitudes  of 
the  cylinders  AX,  EO  are  reciprocally  proportional ;  that  is, 
as  the  base  ABCD  to  the  base  EFGH,  so  is  the  altitude  MN  to 
the  altitude  KL. 

Either  the  altitude  MN  is  equal  to  the  altitude  KL,  or  these 
altitudes  are  not  equal.  First,  let  them  be  equal ;  and  the 
cylinders  AX,  EO  being  also  equal,  and  cones  and  cylinders 
a  11.12.  of  the  same  altitude  being  to  one  another  as  their  basest,  there- 
b  A.  5.  fore  the  base  ABCD  is  equal  ^  to  the  base  EFGH  ;  and  as  the 
base  ABCD  is  to  the  base  EFGH,  so  is  the  altitude  MN  to 
the    altitude    KL. 

But   let   the   alti-  R 

tudes     KL,     MN 
be    unequal,    and 
MN   the   greater 
of  the   two,   and 
from     MN     take 
MP  equal  to  KL, 
and,  through  the 
point  P,  cut  the 
cylinder    EO    by 
the    plane    TVS 
parallel  to  the  op- 
posite planes  of  the  circles  EFGH,  RO ;    therefore  the   com- 
mon section  of  the  plane  TYS  and  the  cylinder  EO  is  a  cir- 
cle, and  consequently  ES  is  a  cylinder,  the  base  of  which  is  the 
circle  EFGH,  and  altitude  MP :   and  because  the  cylinder  AX 
is  equal  to  the  cylinder  EO,  as  AX  is  to  the  cyHnder  ES,  so 
c7.  5.      'is  the  cylinder  EO  to  the  same  ES.     But  as  the  cylinder  A^ 
to  the  cylinder  ES,  so  ^  is  the  base  ABCD  to  the  base  EFGH ; 
for  the  cylinders  AX,  ES  are  of  the  same  altitude ;    and  as  the 
d  13. 12.  cylinder  EO  to  the  cylinder  ES,  so  ^  is  the  altitude  MN  to 
the  altitude  MP,  because  the  cylinder  EO  is  cut  by  the  plane 


OF  EUCLID.  2S2 

TYS  parallel  to  its  opposite  planes.      Therefore,  as  the  base  B.  XII. 
ABCD  to  the  base  EFGH,  so  is  the  altitude  MN  to  the  altitude  ^— y^-* 
MP :    but  MP  is  equal  to  the  altitude  KL ;    wherefore   as  the 
base  ABCD  to  the  base  EFGH,  so  is  the  altitude  MN  to  the  al- 
titude KL ;   that  is,  the  bases  and  altitudes  of  the  equal  cylinders 
AX,  EO  are  reciprocally  proportional. 

But  let  the  bases  and  altitudes  of  the  cylinders  AX,  EO  be 
reciprocally  proportional,  viz.  the  base  ABCD  to  the  base  EFGH, 
as  the  altitude  MN  to  the  altitude  KL :  the  cylinder  AX  is  equal 
to  the  cylinder  EO. 

First,  let  the  base  ABCD  be  equal  to  the  base  EFGH ;   then 
because  as  the  base  ABCD  is  to  the  base  EFGH,  so  is  the  alti- 
tude MN  to  the  altitude  KL  ;  MN  is  equally  to  KL,  and  therefore  b.  A.  5. 
the  cylinder  AX  is  equal  ^  to  the  cylinder  EO.  a  11. 12. 

But  let  the  bases  ABCD,  EFGH  be  unequal,  and  let  ABCD 
be  the  greater ;  and  because  as  ABCD  is  to  the  base  EFGH,  so 
is  the  altitude  MN  to  the  altitude  KL :  therefore  MN  is  great- 
er b  than  KL.  Then,  the  same  construction  being  made  as  be- 
fore, because  as  the  base  ABCD  to  the  base  EFGH,  so  is  the  al- 
titude MN  to  the  altitude  KL :  and  because  the  altitude  KL 
is  equal  to  the  altitude  MP  ;  therefore  the  base  ABCD  is  »  to 
the  base  EFGH,  as  the  cylinder  AX  to  the  cylinder  ES ;  and  a^^ 
the  altitude  MN  to  the  altitude  MP  or  KL,  so  is  the  cylinder  EO 
to  the  cylinder  ES :  therefore  the  cylinder  AX  is  to  the  cylinder 
ES,  as  the  cylinder  EO  to  the  same  ES ;  whence  the  cylinder 
AX  is  equal  to  the  cylinder  EO  :  and  the  same  reasoning  holds 
in  cones.     Q.  E.  D. 


PROP.  XVI.    PROB. 


TO  describe  in  the  greater  of  two  circles  that  have 
the  same  centre,  a  polygon  of  an  even  number  of 
equal  sides,  that  shall  not  meet  the  lesser  circle. 


Let  ABCD,  EFGH  be  two  given  circles  having  the  same  cen- 
tre K :  it  is  required  to  inscribe  in  the  greater  circle  ABCD  a 
polygon  of  an  even  number  of  equal  sides,  that  shall  not  meet 
the  lesser  circle. 

Through  the  centre  K  draw  the  straight  line  BD,  and  from 
the  point  G,  where  it  meets  the  circumference  of  the  lesser 


2S4  THE  ELEMENTS 

B.  XII.  circle,  draw  GA  at  right  angles  to  BD,  and  produce  it  to  C; 
^^"•v--^  therefore  AC  touches  ^  the  circle  EFGH :    then,  if  the  circum- 
a  16.  3.     ference  BAD  be  bisected,  and  the  half  of  it  be  again  bisected, 
b  Lem-    and  so  on,  there  must  at  length  remain  a  circumference  less  ^ 
"^a.         than   AD;    let    this   be    LD ;    and 
from   the  point   L  draw    LM   per- 
pendicular   to    BD,     and    produce 
it  to  N  ;  and  join  LD,  DN.    There- 
fore LD  is  equal  to  DN :    and  be- 
cause LN  is  parallel  to  AC,  and  that 
AC    touches    the     circle     EFGH  ; 
therefore    LN   does    not   meet  the 
circle  EF'GH  :    and  much  less  shall 
the  straight  lines  LD,  DN  meet  the 
circle   EFGH:    so   that  if  straight 
lines  equal  to  LD  be  applied  in  the  circle  ABCD  from  the  point 
L  around  to  N,  there  shall  be  described  in  the  circle  a  polygon 
of  an  even  number  of  equal  sides  not  meeting  the  lesser  circle.. 
Which  was  to  be  done. 


LEMMA  n. 

IF  two  trapeziums  ABCD,  EFGH  be  inscribed 
in  the  circles,  the  centres  of  which  are  the  points  K, 
L ;  and  if  the  sides  AB,  DC  be  parallel,  as  also  EF, 
HG ;  and  the  other  four  sides  AD,  BC,  EH,  FG  be 
all  equal  to  one  another;  but  the  side  AB  greater 
than  EF,  and  DC  greater  than  HG  :  the  straight 
line  KA  from  the  centre  of  the  circle  in  which  the 
greater  sides  are,  is  greater  than  the  straight  line  LE 
drawn  from  the  centre  to  the  circumference  of  the 
other  circle. 

If  it  be  possible,  let  KA  be  not  greater  than  LE ;  then  KA 
must  be  cither  equal  to  it,  or  less.  F'irst,  let  KA  be  equal 
to  LE :  therefore,  because  in  two  equal  circles,  AD,  BC  in  the 
one  are  equal  to  EH,  FG  in  the  other,  the  circumferences 
a2B.  3.  AD,  BC  arc  equal  »  to  the  circumferences  EH,  FG ;  but  be- 
cause the  straight  lines  AB,  DC  are  respectively  greater  than 
EF,  GH,  the  circumferences  AB,  DC  are  greater  than  EF, 
HG  :  therefore  the  whole  circumference  ABCD  is  greater 
than  the   v.hole  EFGH;    but  it  is  also  equctl  to  it,  which  is 


OF  EUCLID. 


28-5 


impossible :   therefore   the    straight   line  KA  is  not   equal   to  B.  Xll. 

But  let  KA  be  less  than  LE,  and  make  LM  equal  to  KA, 
and  from  the  centre  L,  and  distance  LM,  describe  the  circle 
MNOP,  meeting  the  straight  lines  LE,  LF,  LG,  LH,  in  M, 
N,  O,  P  ;  and  join  MN,  NO,  OP,  PM,  which  are  respectively 
parallel  »  to,  and  less  than  EF,  FG,  GH,  HE :  then  because  EH  a  2.  6. 
is  greater  than  MP,  AD  is  greater  than  MP  j  and  the  circles 


>ABCD,  MNOP  are  equal;  therefore  the  circumference  AD  is 
greater  than  MP ;  for  the  same  reason,  the  circumference  BC 
is  greater  than  NO ;  and  because  the  straight  line  AB  is  greater 
than  EF,  which  is  greater  than  MN,  much  more  is  AB  greater 
than  MN  :  therefore  the  circumference  AB  is  greater  than  MN ; 
and,  for  the  same  reason,  the  circumference  DC  is  greater  than 
PO :  therefore  the  whole  circumference  ABCD  is  greater  than 
the  whole  MNOP  ;  but  it  is  likewise  equal  to  it,  which  is  impossi- 
ble :  therefore  K A  is  not  less  than  LE ;  nor  is  it  equal  to  it ; 
the  straight  line  KA  must  therefore  be  greater  than  LE.  Q.  E.  D. 
CoR.  And  if  there  be  an  isosceles  triangle,  the  sides  of  Avhich 
are  equal  to  AD,  BC,  but  its  base  less  than  AB  the  greater  of 
the  two  sides  AB,  DC ;  the  straight  line  KA  may,  in  the  same 
manner,  be  demonstrated  to  be  greater  than  the  straight  line 
drawn  from  the  centre  to  the  circumference  of  the  circle  de- 
scribed about  the  triangle. 


THE  ELEMENTS 


PROP.  XVII.    PROB. 


See  N.        TO  describe  in  the  greater  of  two  spheres  whiclx. 
have  the  same  centre,  a  solid  polyhedron,  the  super- 
ficies of  which  shall  not  meet  the  lesser  sphere. 

Let  there  be  two  spheres  about  the  same  centre  A  ;  it  is  re-: 
quired  to  describe  in  the  greater  a  solid  polyhedron,  the  superfi- 
cies of  which  shall  not  meet  the  lesser  sphere. 

Let  the  spheres  be  cut  by  a  plane  passing  through  the  centre  ; 
the  common  sections  of  it  with  the  spheres  shall  be  circles ; 
because  the  sphere  is  described  by  the  revolution  of  a  semicir- 
cle about  the  diameter  remaining  unmoveable ;  so  that  in  what- 
ever position  the  semicircle  be  conceived,  the  common  section 
of  the  plane  in  which  it  is  with  the  superficies  of  the  sphere  is 
the  circumference  of  a  circle ;  and  this  is  a  great  circle  of  the 
sphere,  because  the  diameter  of  the  sphere,  which  is  likewise 

a  IS.  3.  the  diameter  of  the  circle,  is  greater  *  than  any  straight  line 
in  the  circle  or  sphere :  let  then  the  circle  made  by  the  section 
of  the  plane  with  the  greater  sphere  be  BCDE,  and  with  the 
lesser  sphere  be  FGH  ;  and  draw  the  two  diameters  BD,  CE 
at  right  angles  to  one  another;    and  in  BCDE,  the  greater  of 

b  16. 12.  the  two  circles,  describe  ^  a  polygon  of  an  even  number  of  e- 
qual  sides  not  meeting  the  lesser  circle  FGH :  and  let  its  sides, 
in  BE,  the  fourth  part  of  the  circle,  be  BK,  KL,  LM,  ME  ; 
join  KA  and  produce  it  to  N ;  and  from  A  draw  AX  at  right 
angles  to  the  plane  of  the  circle  BCDE  meeting  the  superficies 
of  the  sphere  in  the  point  X;  and  let  planes  pass  through  AX, 
and  each  of  the  straight  lines  BD,  KN,  which,  from  what 
has  been  said,  shall  produce  great  circles  on  the  superficies  of 
the  sphere,  and  let  BXD,  KXN  be  the  semicircles  thus  made 
upon  the  diameters  BD,  KN :  therefore  because  X A  is  at  right 
angles  to  the  plane  of  the  circle  BCDE,  every  plane  which 

c  18. 11.  passes  through  XA  is  at  right  <=■  angles  to  the  plane  pf  the  circle 
BCDE  ;  wherefore  the  semicircles  BXD,  KXN  are  at  right 
angles  to  that  plane:  and  because  the  semicircles  BED,  BXD, 
KXN,  upon  the  equal  diameters  BD,  KN,  are  equal  to  one 
another,  their  halves  BE,  BX,  KX,  arc  equal  to  one  an- 
other: therefore,  as  many  sides  of  the  polygon  as  are  in  BE, 
so  many  there  are  in  BX,  KX  equal  to  the  sides  BK, 
KL,  LM,  ME:  let  these  polygons  be  described,  and  their 
sides   be   BO,   OP,   PR,   RX ;    KS,   ST,   TY,    YX,    and  join 


OF  EUCLID. 


287 


OS,  PT,  RY ;  and  from  the  points  O,  S,  draw  OV,  SQ  perpen-  b.  Xii. 
diculars  to  AB,  AK :  and  because  the  plane  BOXD  is  at  right  ^-^-ymJ 
angles  to  the  plane  BCDE,   and  in  one  of  them  BOXD,  OV 
is  drawn  perpendicular  to  AB  the  common  section  of  the  planes, 
therefore  O  V  is  perpendicular  a  to  the  plane   BCDE  :    for  the  a  4.  def. 
same  reason  SQ  is  perpendicular  to  the  same  plane,  because   li- 
the plane  KSXN  is  at  right  angles  to  the  plane  BCDE.      Join 
VQ;    and  because  in  the  equal  semicircles  BXD,  KXN  the 


circumferences  BO,  KS  are  equal,  and  OV,  SQ  are  perpen- 
dicular to   their   diameters,   therefore  d  OV   is   equal    to   SQ,  d26.1. 
and  BV  equal  to  KQ:  but  the  whole  BA  is  equal  to  the  whole 
KA,  therefore  the  remainder  VA  is   equal   to  the   remainder 
QA:    as  therefore  BV  is  to  VA,  so  is  KQ  to  QA,  wherefore 
VQ  is  parallel  «  to  BK:    and  because  OV,   SQ   are   each  ofe2. 6. 
them  at  right  angles  to  the  plane  of  the  circle  BCDE,  OV  is 
parallel*"  to  SQ;    and  it  has  been  proved  that  it  is  also   equal  f  6.  It. 
to  it ;  therefore  QV,  SO  are  equal  and  parallel  s  :    and  because  g  33. 1- 
QV  is  parallel  to  SO,  and  also  to  KB  ;    OS  is  parallel  i^  to  BK ;  b  9. 11. 
and  therefore  BO,  KS  which  join  them  are  in  the  same  plane 


2S8 


THE  ELEMENTS 


B.  XII.  in  which  these  parallels  are,  and  the  quadrilateral  figure  KBOS 

V->vJ  is  in  one  plane :  and  if  PB,  TK  be  joined,  and  perpendiculars 
be  drawn  from  the  points  P,  T  to  the  straight  lines  AB,  AK, 
it  may  be  demonstrated,  that  TP  is  parallel  to  KB  in  the  very 
same  way  that  SO  was  shown  to  be  parallel  to  the  same  KB ; 

a  9.  11.  wherefore  *  TP  is  parallel  to  SO,  and  the  quadrilateral  figure 
SOPT  is  in  one  plane :  for  the  same  reason,  the  quadrilateral 

b  2. 11.    TPRY  is  in  one  plane :  and  thp  figure  YRX  is  also  in  one  plane''. 


Therefore,  if  from  the  points  O,  S,  P,  T,  R,  Y  there  be  drawn 
straight  lines  to  the  point  A,  there  shall  be  formed  a  solid  po- 
lyhedron between  the  circumferences  BX,  KX  composed  o£ 
pyramids,  the  bases  of  which  are  the  quadrilaterals  KBOS, 
SOPT,  TPRY,  and  the  triangle  YRX,  and  of  which  the  com- 
mon vertex  is  the  point  A  :  and  if  the  same  construction  be 
made  upon  each  of  the  sides  KL,  LM,  ME,  as  has  been  done 
upon  BK,  and  the  like  be  done  also  in  the  other  three  qua- 
drants, and  in  the  other  hemisphere  ;  there  shall  be  for- 
med   a    solid    polyhedron   described    in    the    sphere,    compo- 


OF  EUCLIt).  289 

sed  of  pyramids,  the  bases  of  which  are  the  aforesaid  quadri-  B.  XII. 
lateral   figures,    and   the   triangle   YRX,  and  those  formed   in  ^->v--^ 
the  like  manner  in  the  rest  of  the  sphere,  the  common  vertex 
of  them  all  being  the  point  A :  and  the  superficies  of  this  so- 
lid polyhedron  does  not  meet  the  lesser  sphere  in  which  is  the 
circle  FGH:   for,  from  the  point  A  draw  *  AZ   perpendicular  a  11.  11. 
to  the  plane  of  the  quadrilateral  KBOS  meeting  it  in  Z,  and 
join   BZ,  ZK :  and  because  AZ  is  perpendicular  to  the  plane 
KBOS,  it  makes  right  angles  with  every  straight  line  meeting 
it  in  that  plane  ;  therefore  AZ  is  perpendicular  to  BZ  and  ZK : 
and  because  AB  is  equal  to  AK,  and  that  the  squares  of  AZ, 
ZB,  are  equal  to  the  square  of  AB;  and  the  squares  of  AZ, 
ZK  to  the  square  of  AK  ^  ;  therefore  the  squares  of  AZ,  ZB  b  4,7- 1- 
are  equal  to  the  squares  of  AZ,  ZK :   take  from  these  equals 
the  square  of  AZ,  the  remaining  square  of  BZ  is  equal  to  the 
I'emaining   square  of  ZK ;  and  therefore  the  straight  line  BZ 
is  equal  to  ZK ;  in  the  Hke  manner  it  may  be  demonstrated, 
that  the  straight  lines  drawn  from  the  point  Z  to  the  points  O, 
S  are  equal  to  BZ  or  ZK  :  therefore  the  circle  described  from 
the  centre  Z,  and  distance  ZB,  shall  pass  through  the  points  K,  O, 
S,  and  KBOS  shall  be  a  quadrilateral  figure  in  the  circle:   and 
because  KB  is  greater  than  QV,  and  QV  equal  to  SO,  there- 
fore KB  is  greater  than  SO :   but  KB  is  equal  to  each  of  the 
straight  lines  BO,  KS ;   wherefore  each  of  the  circumferences 
cut  oft"  by  KB,  BO,  KS  is  greater  than  that  cut  oft"  by  OS ;  and 
these  three  circumferences,  together  with  a  fourth  equal  to  one 
of  them,  are  greater  than  the  sUme  three  together  with  that  cut 
off"  by  OS ;   that  is,  than  the  whole  circumference  of  the  cir- 
cle ;   therefore  the  circumference  subtended  by  KB  is  greater 
than  the  fourth  part  of  the  whole  circumference  of  the  circle 
KBOS,  and  consequently  the  angle  BZK  at  the  centre  is  great- 
er than  a  right  angle :  and  because  the  angle  BZK  is  obtuse, 
the  square  of  BK  is  greater '=  than   the   squares  of  BZ,    ZK;cl2.  2. 
that  is,  greater  than  twice  the  square   of  BZ.     Join  KV,  and 
because  in  the  triangles  KBVj  OBV,  KB,  BV  are  equal  to  OB, 
BV,  and   that  they  contain  equal   angles;   the  angle  KVB  is 
equal  ^  to  the  angle  OVB :  and  OVB  is  a  right  angle ;  there-  d  4. 1. 
fore  also  KVB  is  a  right  angle:  and  because  BD  is  less  than 
twice  DV,  the  rectangle   contained   by   DB,  BV  is  less   than 
twice  the  rectangle  DVB  ;  that  is  e,  the  square  of  KB  is  less  e  8.  6. 
than  twice  the  square  of  KV :  but  the  square  of  KB  is  greater 
than  twice  the  square  of  BZ  :   therefore  the  square  of  KV   is 
greater  than  the  square  of  BZ :   and  because  BA   is  equal  to 
AK,  and  that  the  squares  of  BZ,   ZA  are  equal  together  to  the 
square  of  BA,  and  the  squares  of  KV,  VA  to  the   square  of 
AK;  therefore  the  squares  of  BZ,  ZA  are  equal  to  the  squares 
of  KV,  VA  ;  and  of  these  the  square  of  KV  is  greater  than  the 

2  O 


290  THE   ELEMENTS 

B.  XII.  square  of  BZ ;  therefore  the  square  of  VA  is  less  than  the 
^— •v-^-^  square  of  ZA,  and  the  straight  line  AZ  greater  than  VA  : 
much  more  then  is  AZ  greater  than  AG  ;  because,  in  the  pre- 
ceding proposition,  it  was  shown  that  KV  falls  without  the 
circle  FGH  :  and  AZ  is  perpendicular  to  the  plain  KBOS,  and 
is  therefore  the  shortest  of  all  the  straight  lines  that  can  be  drawn 
from  A,  the  centre  of  the  sphere  to  that  plane.  Therefore  the 
plane  KBOS  does  not  meet  the  lesser  sphere. 

And  that  the  other  planes  bet^veen   the   quadrants  BX,  KX 
fall  without  the  lesser  sphere,  is  thus  demonstrated  :  from  the 
point  A  draw  AI  perpendicular  to   the  plane   of  the   quadrila- 
teral  SOPT,  and  join  lO  ;   and,   as   was   demonstrated  of  the 
plane  KBOS  and  the  point  Z,  in  the  same  way  it  may  be  shown 
that  the  point  I  is  the  centre  of  a  circle  described  about  SOPT  ; 
and   that  OS  is  greater  than  PT  ;   and  PT  was  shown  to  be  pa- 
rallel  to  OS:  therefore,    because  the    two  trapeziums  KBOS, 
SOPT  inscribed  in  circles  have  their  sides  BK,  OS  parallel,  as 
also  OS,   PT  ;  and  their  other  sides  BO,  KS,  OP,  ST  all  equal 
to  one   another,     and  that    BK  is   greater    than  OS,     and  OS 
a  2.  km.  gi-eater  than   PT,  therefore   the   straight  line  ZB    is   greater"* 
^2-         than   lO.     Join  AO   which  will  be  equal  to  AB  ;  and  because 
AIO,  AZB  are  right  angles,  the    squares  of  AI,  lO  are   equal 
to  the  square  of  AO  or  of  AB  ;    that  is,    to  the  squares  of  AZ, 
ZB  ;   and  the  square  of  ZB  is  greater  than  the   square   of  lO, 
therefore  the  square  of  AZ  is  less  than  the  square  of  AI  ;    and 
the  straight  line   AZ  less  than  the  straight  line  AI  :  and  it  was 
proved  that  AZ  is  greater  than  AG  ;    much  more  then  is  AI 
greater  than  AG  ;   therefore  the  plane  SOPT  falls  wholly  with- 
out the  lesser  sphere  :  in   the  same  manner  it  may  be  demon- 
strated that  the  plane  TPRY  falls  without  the  same  sphere,  as 
also  the  triangle   YRX,  viz.  by  the   cor.   of  2d   lemma.     And 
after  the  same  way  it  may  be  demonstrated  that  all  the   planes 
which    contain   the   solid    polyhedron,    fall   without  the   lesser 
sphere.    Therefore  in  the  greater  of  two  spheres  which  have  the 
same   centre,  a  solid  polyhedron  is  described,  the  superficies  of 
which  does  not  meet  the  lesser  sphere.     Which  was  to  be  done. 
But  the  straight  line  AZ  may  be  demonstrated  to  be  greater 
than  AG  otherwise,    and  in  a  shorter  manner,   without  the  help 
of  prop.  16,  as  follows.     From  the  point  G  draw  GU  at  right 
angles  to  AG,  and  join  AU.     If  then  the  circumference  BE  be 
bisected,  and  its  half   again  bisected,   and  so  on,  there  will  at 
length  be   left   a    circumference    less    than    the   circumference 
which  is  subtended  by  a  straight  line  equal  to  GU  inscribed  in 
the  circle  BCDE  :  let  this  be  the  circumference  KB  :  therefore 
the  straight  line  KB  is  less  than   GU  :  and  because  the  angle 
BZK  is  obtuse,  as   was  proved  in  the  preceding,  therefore  BK 
is  greater  than  BZ  :  but  GU  is  greater  than  BK  ;  much  more 


OF  EUCLID.  291 

then  is  GU  greater  than  BZ,  and  the  square  of  GU  than  the  B.  XII. 
square  of  BZ ;  and  AU  is  equal  to  AB  ;  therefore  the  square  ^^»v— J 
of  AU,  that  is,  the  squares  of  AG,  GU,  are  equal  to  the  square 
of  AB,  that  is,  to  the  squares  of  AZ,  ZB  ;  but  the  square  of  BZ 
is  less  than  the  square  of  GU ;  therefore  the  square  of  AZ  is 
greater  than  the  square  of  AG,  and  the  straight  line  AZ  con- 
sequently greater  than  the  straight  line  AG. 

Cor.  And  if  in  the  lesser  sphere  there  be  described  a  solid 
polyhedron  by  drawing  straight  lines  betwixt  the  points  in 
which  the  straight  lines  from  the  centre  of  the  sphere  drawn 
to  all  the  angles  of  the  solid  polyhedron  in  the  greater  sphere 
meet  the  superficies  of  the  lesser ;  in  the  same  order  in  which 
are  joined  the  points  in  which  the  same  lines  from  the  centre 
meet  the  superficies  of  the  greater  sphere ;  the  solid  polyhe- 
dron in  the  sphere  BCDE  has  to  this  other  solid  polyhedron 
the  triplicate  ratio  of  that  which  the  diameter  of  the  sphere 
BCDE  has  to  the  diameter  of  the  other  sphere  :  for  if  these 
two  solids  be  divided  into  the  same  number  of  pyramids,  and 
in  the  same  order,  the  pyramids  shall  be  similar  to  one  ano- 
ther, each  to  each  ;  because  they  have  the  solid  angles  at  their 
common  vertex,  the  centre  of  the  sphere,  the  same  in  each  py- 
ramid, and  their  other  solid  angle  at  the  bases  equal  to  one 
another,  each  to  each  ^,  because  they  are  contained  by  three  a  B.  11. 
plane  angles  equal  each  to  each  :  and  the  pyramids  are  contained 
by  the  same  number  of  similar  planes  ;  and  are  therefore  similar  ^  ^  H- 
to  one  another,  each  to  each :  but  similar  pyramids  have  to  ^^" 
one  another  the  triplicate  '^  ratio  of  their  homologous  sides,  c  Cor.  8. 
Therefore  the  pyramid  of  which  the  base  is  the  quadrilateral  12. 
KBOS,  and  vertex  A,  has  to  the  pyramid  in  the  other  sphere 
of  the  same  order,  the  triplicate  ratio  of  their  homologous 
sides ;  that  is,  of  that  ratio  which  AB  from  the  centre  of  the 
greater  sphere  has  to  the  straight  line  from  the  same  centre  to 
the  superficies  of  the  lesser  sphere.  And  in  like  manner,  each 
pyramid  in  the  greater  sphere  has  to  each  of  the  same  order  in 
the  lesser,  the  triplicate  ratio  of  that  which  AB  has  to  the  se- 
midiameter  of  the  lesser  sphere.  And  as  one  antecedent  is  to 
its  consequent,  so  are  all  the  antecedents  to  all  the  consequents. 
Wherefore  the  whole  solid  polyhedron  in  the  greater  sphere  has 
to  the  whole  solid  polyhedron  in  the  other,  the  triplicate  ratio 
of  that  which  AB  the  semidiameter  of  the  first  has  to  the  semi- 
diameter  of  the  other ;  that  is,  which  the  diameter  BD  of  the 
greater  has  to  the  diameter  of  the  other  sphere^ 


a  ir.  12. 


THE  ELEMENTS 


PROP.  XVIII.     THEOR. 

SPHERES  have  to  one  another  the  triplicate  ra- 
tio of  that  which  their  diameters  have. 

Let  ABC,  DEF  be  two  spheres,  of  which  the  diameters  are 
BC,  EF.  The  sphere  ABC  has  to  the  spliere  DEF  the  triplicate 
ratio  of  that  which  BC  has  to  EF. 

For,  if  it  has  not,  the  sphere  ABC  shall  have  to  a  sphere  ei- 
ther less  or  greater  than  UEF,  the  triplicate  ratio  of  that  which 
BC  has  to  EF.  First,  let  it  have  that  ratio  to  a  less,  viz. 
to  the  sphere  GHK  ;  and  let  the  sphere  DEF  have  the  same 
centre  with  GHK ;    and  in  the  greater  sphere  DEF  describe  ^ 


bCor 
17.  12. 


c  14.  5. 


a  solid  polyhedron,  the  superficies  of  which  does  not  meet  the 
lesser  sphere  GHK ;  and  in  the  sphere  ABC  describe  another 
similar  to  that  in  the  sphere  DEF:  therefore  the  solid  polyhe- 
dron in  the  sphere  ABC  has  to  the  solid  polyhedron  in  the 
sphere  DEF,  the  triplicate  ratio  ^  of  that  which  BC  has  to  EF. 
Bqt  the  sphere  ABC  has  to  the  sphere  GHK  the  triplicate  ra- 
tio of  that  which  BC  has  to  EF ;  therefore,  as  the  sphere  ABC 
to  the  sphere  GHK,  so  is  the  solid  polyhedron  in  the  sphere  ABC 
to  the  solid  polyhedron  in  the  sphere  DEF :  but  the  sphere 
ABC  is  greater  than  the  solid  polyhedron  in  it;  therefore  ^  al- 
so the  sphere  GHK  is  greater  than  the  solid  polyhedron  in  the 
sphere  DEF:  but  it  is  also  less,  because  it  is  contained  within 
it,  which  is  impossible :  therefore  the  sphere  ABC  has  not  to 
any  sphere  less  than  DEF  the  triplicate  ratio  of  that  which 
BC  has  to  EF.  In  the  same  manner  it  may  be  demonstrated, 
that  the  sphere  DEF  has  not  to  any  sphere  less  than  ABC  the 
triplicate  ratio  of  that  which  EF  has  to  BC.  Nor  can  the 
sphere  ABC'  have  to  any  sphere  greater  than  DEF,  the  tripli- 
cate ratio  of  that  which  BC  has  to  EF:    for,  if  it  can,  let  it 


OF  EUCLID.  293 

have  that  ratio  to  a  greater  sphere  LMN:  therefore,  by  inver-  B.  XII. 
sion,  the  sphere  LMN  has  to  the  sphere  ABC,  the  tripHcate  ^^\^-J 
ratio  of  that  which  the  diameter  EF  has  to  the  diameter  BC. 
But,  as  the  sphere  LMN  to  ABC,  so  is  the  sphere  DEF  to  some 
sphere,  which  must  be  less  <=  than  the  sphere  ABC,  because  the  c  14.  S, 
sphere  LMN  is  greater  than  the  sphere  DEF :   therefore  the 
sphere  DEF  has  to  a  sphere  less  than  ABC  the  triplicate  ratio 
of  that  which  EF  has  to  BC ;  which  was  shown  to  be  impos- 
sible :  therefore  the  sphere  ABC  has  not  to  any  sphere  greater 
than  DEF  the  triplicate  ratio  of  that  which  BC  has  to  EF :  and 
it  was  demonstrated,  that  neither  has  it  that  ratio  to  any  sphere 
less  than  DEF.     Therefore  the  sphere  ABC  has  to  the  sphere 
DEF,  the  triplicate  ratio  of  that  which  BC  has  to  EF.  Q.  E.  Do 


FINIS. 


NOTES, 

CRITICAL  AND  GEOMETRICAL  : 

coirrAiNiNC 

AN  ACCOUNT  OF  THOSE  THINGS  IN  WHICH  THIS  EDITION 
DIFFERS  FROM  THE  GREEK  TEXT, 

AND 

THE  REASONS  OF  THE  ALTERATIONS  WHICH  HAVE  BEEN  MADE, 

AS    ALSO 

OBSERVATIONS 
ON  SOME  OF  THE  PROPOSITIONS. 


BY  ROBERT  SIMSOJV,  M.  D. 

EMEKITUS  PROFESSOR  OF  MATHEMATICS  IN  THE  UNIVERSITY 
OF    GLASGOW. 


PHILADELPHIA, 

PRINTED    BY    THOMAS    AND    GEORGE    PALMER, 
116,    HIGH    STREET. 

1806. 


NOTES,  &c. 


DEFINITION  I.     BOOK  I. 

It  is  necessary  to  consider  a  solid,  that  is,  a  magnitude  which 
has  length,  breadth,  and  thickness,  in  order  to  understand  aright 
the  definitions  of  a  point,  line,  and  superficies  ;  for  these  all 
arise  from  a  solid,  and  exist  in  it :  the  boundary,  or  bounda- 
lies,  which  contain  a  solid  are  called  superficies,  or  the  boun- 
dary which  is  common  to  two  solids  which  are  contiguous,  or 
which  divides  one  solid  into  two  contiguous  parts,  is  called   a 
superficies :  thus,  if  BCGF  be  one  of  the  boundaries  which 
contain  the  solid  ABCDEFGH,  or  which  is  the  common  boun- 
dary of  this  solid,  and  the  solid  BKLCFNMG,  and  is  there- 
fore in  the  one  as  well  as  in  the  other  solid,  called  a  superficies, 
and  has  no  thickness  :  for  if  it  have  any,  this  thickness  must 
cither  be  a  part  of  the  thickness 
of  the    solid    AG,    or    the  solid 
BM,  or  a  part  of  the  thickness  of  E 
each  of  them.     It   cannot    be    a 
part  of  the  thickness  of  the  solid 
BM ;  because  if  this  solid  be  re- 
moved from  the    solid    AG,    the 
superficies  BCGF,  the   boundary 
of  the  solid  AG  remains  still  the 
same  as  it  was.     Nor  can  it  be  a 
part  of  the  thickness  of  the  solid  AG  ;  because,  if  this  be  re- 
moved from  the  solid  BM,  the  superficies  BCGF,  the  boundary 
of  the  solid  BM,  does  nevertheless  remain:  therefore  the  super- 
ficies BCGF  has  no  thickness,  but  only  length  and  breadth. 

The  boundary  of  a  superficies  is  called  a  line,  or  a  line  is  the 
common  boundary  of  tAvo  superficies  that  are  contiguous,  or 
which  divides  one  superficies  into  two  contiguous  parts  :  thus, 
if  R  '  be  one  of  the  boundaries  which  contain  the  superficies 
ABCD,  or  which  is  the  common  boundary  of  this  superficies, 
and  of  the  superficies  KBCL  which  is  contiguous  to  it,  this 
boundary  BC  is  called  a  line,  and  has  no  breadth  :  for  if  it 
have  any,  this  must  be  part  either  of  the  breadth  of  the  super^ 
ficies  ABCD,  or  of  the  superficies  KBCL,  or  part  of  each  of 

3  P 


298 


NOTES. 


H 


M 


D 


Book  I.  them.  It  is  not  the  part  of  the  breadth  of  the  superficies  KBCL; 

-•'~^^'>»-'  for,  if  this  superficies  be  removed  from  the  superficies  ABCD, 
the  line  BC,  which  is  the  boundary  of  the  superficies  ABCD, 
remains  the  same  as  it  was  :  nor  can  the  breadth  that  BC  is  sup- 
posed to  have  be  a  part  of  the  breadth  of  the  superficies  ABCD; 
because,  if  this  be  removed  from  the  superficies  KBCL,  the  line 
BC,  which  is  the  boundary  of  the  superficies  KBCL,  does  never- 
theless remain :  therefore  the  line  BC  has  no  breadth :  and  be- 
cause the  line  BC  is  a  superficies,  and  that  a  superficies  has  no 
thickness,  as  was  shown  ;  therefore  a  line  has  neither  breadth 
nor  thickness,  but  only  length. 

The  boundary  of  a  line  is  called  a  point,  or  a  point  is  the 
common  boundary  or  extremity 
of  two  lines  that  are  contiguous  : 
thus,  if  B  be  the  extremity  of  the 
line  AB,  or  the  common  extremi- 
ty of  the  two  lines  AB,  KB,  this 
extremity  is  called  a  point,  and 
has  no  length :  for,  if  it  have 
any,  this  length  must  either  be 
part  of  the  length  of  the  line  AB, 
or  of  the  line  KB.     It  is  not  part       A  B  K 

of  the  length  of  KB  ;  for  if  the  line  KB  be  removed  from  AB, 
the  point  B,  which  is  the  extremity  of  the  line  AB,  remains  the 
same  as  it  was  :  nor  is  it  part  of  the  length  of  the  line  AB ;  for, 
if  AB  be  removed  from  the  line  KB,  the  point  B,  which  is  the 
extremity  of  the  line  KB,  does  nevertheless  remain  :  there- 
fore the  point  B  has  no  length :  and  because  a  point  is  in  a  line, 
and  a  line  has  neither  breadth  nor  thickness,  therefore  a  point 
has  no  length,  breadth,  nor  thickness.  And  in  this  manner  the 
definitions  of  a  point,  line,  and  superficies  are  to  be  understood. 


DEF.  VII.     B.  I. 

Instead  of  this  definition  as  it  is  in  the  Greek  copies,  a  more 
distinct  one  is  given  from  a  property  of  a  plane  superficies, 
which  is  manifestly  supposed  in  the  Elements,  viz.  that  a 
straight  line  drawn  from  any  point  in  a  plane  to  any  other  in  it. 
is  wholly  in  that  plane. 

DEF.  VIII.     B.  I. 

It  seems  that  he  who  made  this  definition  designed  that  it 
should  comprehend  not  only  a  plane  angle  contained  by  two 
straight  lines,  but  likewise  the  angle  which  some  conceive  to 
be  made  by  a  straight  line  and  a  curve,  or  by  two  curve  lines, 
w  hich  meet  one  another  in  a  plane :  but  though  the  meaning  of 


NOTES.  299 

the  words  i-x  tuB-etetiy  that  is,  in  a  straight  line,  or  in  the  same  Book  I. 
direction,  be  plain,  when  two  straight  lines  are  said  to  be  in  a  ^*^"^^">w' 
straight  line,  it  does  not  appear  what  ought  to  be  vuiderstood 
by  these  words,  when  a  straight  line  and  a  curve,  or  two  curve 
lines,  are  said  to  be  in  the  same  direction  ;  at  least  it  cannot  be 
explained  in  this  place  ;  which  makes  it  probable  that  this  defi- 
nition, and  that  of  the  angle  of  a  segment,  and  what  is  said  of 
the  angle  of  a  semicircle,  and  the  angles  of  segments,  in  the 
16th  and  31st  propositions  of  book  3,  are  the  additions  of  some 
less  skilful  editor  :  on  which  account,  especially  since  they  are 
quite  useless,  these  definitions  are  distinguished  from  the  rest 
by  inverted  double  commas. 


DEF.  XVII.     B.  I. 

The  words,  "  which  also  divides  the  circle  into  two  equal 
"  parts,"  are  added  at  the  end  of  this  definition  in  all  the  copies, 
but  are  now  left  out  as  not  belonging  to  the  definition,  being 
only  a  corollary  from  it.  Proclus  demonstrates  it  by  conceiving 
one  of  the  parts  into  Avhich  the  diameter  diAides  the  circle,  to 
be  applied  to  the  other  ;  for  it  is  plain  they  must  coincide,  else 
the  straight  lines  from  the  centre  to  the  circumference  would 
not  be  all  equal :  the  same  thing  is  easily  deduced  from  the  3 1st 
prop,  of  book  3,  and  the  24th  of  the  same;  from  the  first  of  which 
it  follows  that  semicircles  are  similar  segments  of  a  circle :  and 
from  the  other,  that  they  are  equal  to  one  another. 


DEF.  XXXIII.     B.  I. 

This  definition  has  one  condition  more  than  is  necessary ;  be- 
cause every  quadrilateral  figure  which  has  its  opposite  sides 
equal  to  one  another,  has  likewise  its  opposite  angles  equal*; 
and  on  the  contrary. 

Let  ABCD  be  a  quadrilateral  figure  of  which  the  opposite 
sides  AB,  CD  are  equal  to  one  another ; 
as  also  AD  and  BC  :  join  BD  ;  the  two 
sides  AD,  DB  are  equal  to  the  two  CB, 
BD,  and  the  base  AB  is  equal  to  the 
base  CD  ;  therefore  by  prop.  8,  of  book 
1,  the  angle  ADB  is  equal  to  the  angle  B  C 

CBD  ;  and  by  prop.  4,  B.  1,  the  angle  BAD  is  equal  to  the 
angle  DCB,  and  ABD  to  BDC  ;  and  therefore  also  the  angle 
ADC  is  equal  to  the  angle  ABC. 


aOQ  NOTES. 

Book  I.        And  if  the  angle  BAD  be  equal  to  the  opposite  angle  BClD, 
^-^f"-^-^*'^  and  the  angle  ABC  to  ADC  ;  the  opposite  sides  are  equal ;  be- 
cause, by  prop.  32,  book  1,  all  the  angles  of  the  quadrilateral 
figure  AHCD  ai^e  together  equal  to  a  D 

four  right  angles,  and  the  two  angles 
BAD,  ADC  are  together  equal  to 
the  two  angles  BCD,  ABC  :  where- 
fore BAD,  ADC  are  the  half  of  all 
the  four   angles  ;  that  is,  BAD  and     B  C 

ADC  are  equal  to  two  right  angles:  and  therefore  AB,  CD  are 
parallels  by  prop.  28,  B.  1.  In  the  same  manner  AD,  BC  are 
parallels  :  therefore  ABCD  is  a  parallelogram,  and  its  opposite 
sides  are  equal  by  prop.  34,  book  I, 


PROP.  VII.     B.  I. 

There  are  two  cases  of  this  proposition,  one  of  which  is  not 
in  the  Greek  text,  but  is  as  necessary  as  the  other :  and  that 
the  case  left  out  has  been  formerly  in  the  text,  appears  plainly 
from  this,  that  the  second  part  of  prop.  5,  which  is  necessary 
to  the  demonstration  of  this  case,  can  be  of  no  use  at  all  in  the 
Elements,  or  anywhere  else,  but  in  this  demonstration ;  because 
the  second  part  of  prop.  5  clearly  follows  from  the  first  part, 
and  prop.  1 3,  book  1 .  This  part  must  therefore  have  been  added 
to  prop.  5,  upon  account  of  some  proposition  betwixt  the  5th 
and  13th,  but  none  of  these  stand  in  need  of  it  except  the  7th 
proposition,  on  account  of  which  it  has  been  added  :  besides, 
the  translation  from  the  Arabic  has  this  case  explicitly  demon- 
strated. And  Proclus  acknowledges,  that  the  second  part  of 
prop.  5  was  added  upon  account  of  prop.  7,  but  gives  a  ridicu- 
lous reason  for  it,  "  that  it  might  afford  an  answer  to  objections 
'•  made  against  the  7th,"  as  if  the  case  of  the  7th,  Avhich  is  left 
out,  were,  as  he  expressly  makes  it,  an  objection  against  the 
proposition  itself.  Whoever  is  curious,  may  read  what  Proclus 
says  ,of  this  in  his  conmientary  on  the  5th  and  7th  propositions ; 
for  it  is  not  Morth  while  to  relate  his  trifles  at  full  length. 

It  was  thought  proper  to  change  the  enunciation  of  this  7th 
prop,  so  as  to  preserve  the  very  same  meaning  ;  the  literal 
translation  from  the  Greek  being  extremely  harsh,  and  difficult 
to  be  understood  by  beginners. 


NOTES. 


PROP.  XI.     B.  I. 


A  corollary  is  added  to  this  proposition,  which  is  necessary  to 
Prop.  1,  B.  11,  and  otherwise. 


PROP.  XX.  and  XXI.     B.  I. 

Proclus,  in  his  commentary,  relates,  that  the  Epicureans  de- 
rided this  proposition,  as  being  manifest  even  to  asses,  and  need- 
ing no  demonstration  ;  and  his  answer  is,  that  though  the  truth 
of  it  be  manifest  to  our  senses,  yet  it  is  science  which  must  give 
the  reason  why  two  sides  of  a  triangle  are  greater  than  the  third : 
but  the  right  answer  to  this  objection  against  this  and  the  2 1st, 
and  some  other  plain  propositions,  is,  that  the  number  of  axioms 
ought  not  to  be  increased  without  necessity,  as  it  must  be  if  these 
propositions  be  not  demonstrated,  Mons.  Clairault,  in  the  pre- 
face to  his  Elements  of  Geometry,  published  in  French  at  Paris, 
ann.  1741,  says,  That  Euclid  has  been  at  the  pains  to  prove, 
that  the  two  sides  of  a  triangle  which  is  included  within  another 
are  together  less  than  the  two  sides  of  the  triangle  which  in- 
cludes it ;  but  he  has  forgot  to  add  this  condition,  viz.  that  the 
triangles  must  be  upon  the  saine  base ;  because,  unless  this  be 
added,  the  sides  of  the  included  triangle  may  be  greater  than  the 
sides  of  the  triangle  which  includes  it,  in  any  ratio  which  is  less 
than  that  of  two  to  one,  as  Pappus  Alexandrinus  has  demon- 
strated in  prop.  3.  B.  3.  of  his  mathematical  collections. 


PROP.  XXII.     B.  I. 

Some  authors  blame  Euclid,  because  he  does  not  demonstrate 
that  the  two  circles  made  use  of  in  the  construction  of  this 
problem  must  cut  one  another  :  but  this  is  very  plain  from  the 
determination  he  has  given,  viz.  that  any  two  of  the  straight 
lines  DF,  FG,  GH  must  be  great- 
er than  the  third:  for  who  is  so 
dull,  though  only  beginning  to 
learn  the  Elements,  as  not  to  per- 
ceive that  the  circle  described 
from  the  centre  F,  at  the  distance 
FD,  must  meet  FH  betwixt  F  and 
H,  because  FD  is  less  than  FH  ;^  ^^  F    G  H 

and  that,  for  the  like  reason,  the  circle  described  from  the  centre 
G,  at  the  distance  GH,  p^'  GM,  must  meet  DG  betwixt  D 


502 


NOTES. 


Book  I.  and  G  ;  and  that  these  circles  must  meet  one  another,  because 

'-«^^^'''^*«*^  FD  and  GH  are  together  greater 
than  FG  ?  And  this  determina- 
tion is  easier  to  be  understood 
than  that  which  Mr.  Thomas 
Simpson  derives  from  it,  and  puts 
instead  of  Euclid's,  in  the  49th 
page  of  his  Elements  of  Geome-  DM  F  G  H 

try,  that  he  may  supply  the  omission  he  blames  Euclid  for  ; 
which  determination  is,  that  any  of  the  three  straight  lines  must 
be  less  than  the  sum,  but  greater  than  the  difference  of  the  other 
two :  from  this  he  shows  the  circles  must  meet  one  another,  in 
one  case  ;  and  says,  that  it  may  be  proved  after  the  same  man- 
ner in  any  other  case  :  but  the  straight  line  GM,  which  he  bids 
take  from  GF,  may  be  greater  than  it,  as  in  the  figure  here  an- 
nexed ;  in  which  case  his  demonstration  must  be  changed  into 
another. 


PROP.  XXIV.     B.  I. 

To  this  is  added,  "  of  the  two  sides  DE,  DF,  let  DE  be 
"  that  which  is  not  greater  than  the  other  ;"  that  is,  take  that 
side  of  the  two  DE,  DF  which  is  not  greater  than  the  other,  in 
order  to  make  with  it  the  angle  EDG  D 
equal  to  BAG  ;  because,  without  this 
restriction,  there  might  be  three  differ- 
ent cases  of  the  proposition,  as  Campa- 
nus  and  others  make. 

Mr.  Thomas  Simpson,  in  p.  262  of 
the  second  edition  of  his  Elements  of 
Geometry,  printed  anno  1760,  observes 
in  his  notes,  that  it  ought  to  have  been 
shown  that  the  point  F  falls  below  the 
line  EG  ;  this  probably  Euclid  omitted 
as  it  is  very  easy  to  perceive,  that  DG 
being  equal  to  DF,  the  point  G  is  in  the  circumference  of  a  cir- 
cle described  from  the  centre  D,  at  the  distance  DF,  and  must 
be  in  that  part  of  it  which  is  above  the  straight  line  EV,  because 
DG  falls  above  DF,  the  angle  EDG  being  greater  than  the  an- 
-Ic  EDF. 


PROP.  XXIX.     B.  I. 

The  proposition  which  is  usually  called  the  5t1i  postulate,  or 
1  Ith  axiom,  bv  some  the  12th,  on  which  this  29th  depends,  has 


NOTES.  303 

given  a  great  deal  to  do,  both  to  ancient  and  modern  geome-  Book  I. 
ters  :  it  seems  not  to  be  properly  placed  among  the  axioms,  as,  '-"^"^^"'^ 
indeed,  it  is  not  self-evident ;  but  it  may  be  demonstrated :  thus 

DEFINITION  1. 

The  distance  of  a  point  from  a  straight  line,  is  the  perpendi- 
cular drawn  to  it  from  the  point. 

DEF.  2. 

One  straight  line  is  said  to  go  nearer  to,  or  further  from, 
another  straight  line,  when  the  distance  of  the  points  of  the 
first  from  the  other  straight  line  become  less  or  greater  than 
they  were  ;  and  two  straight  lines  are  said  to  keep  the  same 
distance  from  one  another,  when  the  distance  of  the  points  of 
one  of  them  from  the  other  is  always  the  same. 

AXIOM. 

A  straight  line  cannot  first  come  nearer  to  another  straight 
line,  and  then  go  further  from  it,  before    A  t) 

it  cuts  it ;  and  in  like  manner,  a  straight    "~^ 

line  cannot  go  further  from  anotherD- E 

straight  line,  and  then  come  nearer  to  F  -~-..,.__— ■ — ^  H 

it ;  nor  can  a  straight  line  keep  the  G 

same  distance  from  another  straight  line,  and  then  come  nearer 
to  it,  or  go  further  from  it ;  for  a  straight  line  keeps  always  the 
same  direction. 

For  example,  the  straight  line  ABC  cannot  first  come  nearer 
to  the  straight  line  DE,  as  from  the  B 

point  A  to  the  point  B,  and  then,    A  -~^  p  q.      . * 

from  the  point  B  to  the  point  C,  go    D  P  figure 

further  from  the  same  DE  :  and  in     F tT H  above. 

like  manner,  the  straight  line  FGH 

cannot  go  further  from  DE,  as  from  F  to  G,  and  then,  from  G 
to  H,  come  nearer  to  the  same  DE  :  and  so  in  the  last  case, 
as  in  fig.  2. 

PROP.  1. 

If  two  equal  straight  lines  AC,  BD,  be  each  at  right  angles 
to  the  same  straight  line  AB  ;  if  the  points  C,  D  be  joined  by 
the  straight  line  CD,  the  straight  line  EF  drawn  from  any 
point  E  in  AB  unto  CD,  at  right  angles  to  AB,  shall  be  equal 
to  AC,  or  BD. 

If  EF  be  not  equal  to  AC,  one  of  them  must  be  greater 
than  tlie  other  ;  let  AC  be  the  greater  ;  then,  because  FE  is 


J04 


NOTES. 


Book  I.  less  than  CA,  the  straight  line  CFD  is  nearer  loathe  straight 
s^'V"^  line  AB  at  the  point  F  than  at  the 

point  C,  that  is,  CF  comes  nearer 

to  AB  from  the  point  C  to  F  :  but 

because  DB  is  greater  than  FE, 

the  straight  line  CFD  is  further 

from  AB  at  the  point  D  than  at  F, 

that  is,  FD  goes  further  from  AB 

from  F  to  D :  therefore  the  straight 

line  CFD  first  comes  nearer  to  the  straight  line  AB,  and  then 

goes  further  from  it,  before  it  cuts  it ;  which  is  impossible.  If 

FEbe  said  to  be  greater  than  C  A,  or  DB,  the  straight  line  CFD 

first  goes  further  from  the  straight  line  AB,  and  then  comes 

nearer  to  it  ;  which  is  also  impossible.      Therefore  FE  is  not 

unequal  to  AC,  that  is,  it  is  equal  to  it. 


PROP. 


C  ^ 


If  two  equal  straight  lines  AC,  BD  be  each  at  right  angles  to 
the  same  straight  line  AB  ;  the  straight  line  CD  which  joins 
their  extremities  makes  right  angles  with  AC  and  BD. 

Join  AD,  BC  ;  and  because,  in  the  triangles  CAB,  DBA, 
CA,  AB  are  equal  to  DB,  BA,  and  the  angle  CAB  equal  to 

a  4  1  the  angle  DBA  ;  the  base  BC  is  equal  a  to  the  base  AD  :  and 
in  the  triangles  ACD,  BDC,  AC,  CD  are  equal  to  BD,  DC, 
and  the  base  AD  is  equal  to  the  base 
BC :    therefore  the  angle    ACD    is 

b  8.  1.  equalhto  the  angle  BDC:  from  any 
point  E  in  AB  draw  EF  unto  CD, 
at  right  angles  to  AB  :  therefore,  by 
prop.  1,  EF  is  equal  to  AC,  or  BD  ; 
wherefore,  as  has  been  just  now 
shown,  the  angle  ACF  is  equal  to  the  angle  EEC  :  in  the  same 
manner,  the  angle  BDF  is  equal  to  tlie  angle  EFD  ;  but  the  an- 
gles ACD,  BDC  are  equal ;  therefore  the  angles  EEC  and  EFD 

c  10. clef.  1.  '^J'e  equal,  and  right  angles^;  wherefore  also  the  angles  ACD, 
BDC  are  right  angles. 

CoR.  Hence,  if  two  straight  lines  AB,  CD  be  at  right  angles 
to  the  same  straight  line  AC,  and  if  betwixt  them  a  straight 
line  BD  be  drawn  at  right  angles  to  either  of  them,  as  to  AB  ; 
then  BD  is  equal  to  AC,  and  BDC  is  a  right  angle. 

if  AC  be  not  equal  to  BD,  take  BG  equal  to  AC,  and 
join  CO :  therefore  by  this  proposition,  the  angle  ACG  is  a 
right  angle  ;  but  ACD  is  also  a  right  angle  ;  wherefore  the  an- 


NOTES. 


^6i 


gles  ACD,  ACG  are  equal  to  one  another,  which  is  impossible.   Book  I. 
Therefore  BD  is  equal  to  AC ;  and  by  this  proposition  BDC  is  v^"v^s» 

a  right  angle. 


PROP.  3. 

If  two  straight  lines  which  contain  ah  angle  be  produced, 
there  may  be  found  in  either  of  them  a  point  from  which  the 
perpendicular  drawn  to  the  other  shall  be  greater  than  any 
given  straight  line. 

Let  AB,  AC  be  two  straight  lines  which  make  an  angle  with 
one  another,  and  let  AD  be  the  given  straight  line ;  a  point  may 
be  found  either  in  AB  or  AC,  as  in  AC,  from  which  the  per- 
pendicular drawn  to  the  other  AB  shall  be  greater  than  AD. 

In  AC  take  any  point  E,  and  draw  ET  perpendicular  to  AB; 
produce  AE  to  G,  so  that  EG  be  equal  to  AE  ;  and  produce 
FE  to  H,  and  make  EH  equal  to  FE,  and  join  HG.  Because, 
in  the  triangles  AEF,  GEH,  AE,  EF  are  equal  to  GE,  EH, 
each  to  each,  and  contain  equal  a  angles,  the  angle  GHE  is  a  15.  1, 
therefore  equal b  to  the  angle  AFE  which  is  a  right  angle :  b  4.  1. 
draw  GK  perpendicular  to  AB  ;  and  because  the  straight  lines 
FK,  HG  are  at  right 

angles  to  FH,  and       A     ,         F  K  B  M 

KG  at  right  angles 
to  FK ;  KG  is  equal  N-  - 
to  FH,  by  cor.  pr.  2, 
that  is,  to  the  double 
of  FE.  In  the  same 
manner,  if  AG  be 
produced  to  L,    so 

that  GL  be  equal  to  AG,  and  LM  be  drawn  perpendicular  to 
AB,  then  LM  is  double  of  GK,  and  so  on.  In  AD  take  AN 
equal  to  FE,  and  AO  equal  to  KG,  that  is,  to  the  double  of  FE, 
or  AN  ;  also,  take  AP  equal  to  LM,  that  is,  to  the  double  of 
KG,  or  AO  ;  and  let  this  be  done  till  the  straight  line  taken  be 
greater  than  AD  :  let  this  straight  line  so  taken  be  AP,  and 
because  AP  is  equal  to  LM,  therefore  LM  is  greater  than  AD, 
Which  was  to  be  done. 


PROP.  4. 

If  two  straight  lines  AB,  CD  make  equal  angles  EAB,  ECD 
with  another  straight  line  EAC  towards  the  same  parts  of  it  5 
AB  and  CD  are  at  right  angles  to  some  straight  line, 

3    O 


306 


NOTES. 


a  15.  1. 

b  4.  1. 


Book  I.  Bisect  AC  in  F,  and  draw  FG  perpendicular  to  AB  ;  take 
^-*•■'^''^"*^  CH  in  the  straight  line  CD  equal  to  AG,  and  on  the  contrary- 
side  of  AC  to  that  on  which  AG  is,  and  join  FH  :  therefore, 
in  the  triangles  AFG,  CFH,  the  sides  FA,  AG  are  equal  to 
FC,  CH,  each  to  each,  and  the  angle 
FAG,  that  a  is,  EAB  is  equal  to  the 
angle  FCH  ;  wherefore  b  the  angle 
AGF  is  equal  to  CHF,  and  AFG  to 
the  angle  CFH :  to  these  last  add  the 
common  angle  AFH  ;  therefore  the 
two  angles  AFG,  AFH  are  equal  to 
the  two  angles  CFH,  HFA,  which 
two  last  are  equal   together  to  two 

right  angles  c,  therefore  also  AFG,  C      H  D 

AFH  are  equal  to  two  right  angles,  and  consequently  d  GF  and 
FH  are  in  one  straight  line.  And  because  AGF  is  a  right  an- 
gle, CHF  which  is  equal  to  it  is  also  a  right  angle  ;  therefore 
the  straight  lines  AB,  CD  are  at  right  angles  to  GH. 


c  13. 
d  14. 


PROP.  5, 


a  23.  1. 


b  13.  1. 


If  two  straight  lines  AB,  CD  be  cut  by  a  third  ACE  so  as  to 
make  the  interior  angles  BAC,  ACD,  on  the  same  side  of  it, 
together  less  than  two  right  angles  ;  AB  and  CD  being  pro- 
duced shall  meet  one  another  towards  the  parts  on  which  are 
the  two  angles  which  are  less  than  two  right  angles. 

At  the  point  C  in  the  sti'aight  line  CE  make  a  the  angle 
ECF  equal  to  the  angle  EAB,  and  draw  to  AB  the  straight 
line  CG  at  right  angles  to  CF  :  then,  because  the  angles  ECF, 
EAB  are  equal  to  one  an-  p 

other,  and  that  the  angles 
ECF,  FCA  are  together 
equal  b  to  two  right  an- 
gles, the  angles  EAB, 
FCA  are  equal  to  two 
right  angles.  But,  by  the 
hypothesis,  the  angles 
EAB,  ACD  are  together 
less  than  two  right  an- 
gles ;  therefore  the  angle 
FCA  is  greater  than 
ACD  and  CD  falls  between  CF  and  AB:  and  because  CF  and 
CD  make  an  angle  with  one  another,  by  prop.  3  a  point  may 
be  found  in  either  of  them  CD,  from  which  the  perpendicular 
drawn  to  CF  shall  be  greater  than  the  straight  line  CG,     Let 


M 


K 


/ 

/ 

X 

\D 

N    ^ 

A    C 

)      ( 

J 

B 

H 


•      NOTES.  307 

this  point  be  H,'  and  draw  HK  perpendicular  to  CF,  meeting  Book  I. 
AB  in  L :  and  because  AB,  CF  contain  equal  angles  with  AC  v-.*'~v'^»w' 
on  the  same  side  of  it  by  prop.  4^  AB  and  CF  are  at  right  angles 
to  the  straight  line  MNO  which  bisects  AC  in  N  and  is  perpen- 
dicular to  CF :  therefore,  by  cor.  prop.  2,  CG  and  KL  which  are 
at  right  angles  to  CF  are  equal  to  one  another :  and  HK  is 
greater  than  CG,  and  therefore  is  greater  than  KL,  and  conse- 
quently the  point  H  is  in  KL  produced.  Wherefore  the  straight 
line  CDH  drawn  betwixt  the  points  C,  H,  which  are  on  contrary 
sides  of  AL,  must  necessarily  cut  the  straight  line  AL. 

PROP.  XXXV.    B.  L 

The  demonstration  of  this  proposition  is  changed,  because,  if 
the  method  which  is  used  in  it  was  followed,  there  would  be 
three  cases  to  be  separately  demonstrated,  as  is  done  in  the 
translation  from  the  Arabic ;  for,  in  the  Elements,  no  case  of  a 
proposition  that  requires  a  different  demonstration,  ought  to  be 
omitted.  On  this  account,  we  have  chosen  the  method  which 
Mons/  Clairault  has  given,  the  first  of  any,  as  far  as  I  know,  in 
his  Elements,  page  21,  and  which  afterward  Mr.  Simpson  gives 
in  his  page  32.  But  whereas  Mr.  Simpson  makes  use  of  prop. 
26,  b.  1,  from  which  the  equality  of  the  two  ti-iangles  does  not 
immediately  follow,  because,  to  prove  that,  the  4  of  b.  1,  must 
likewise  be  made  use  of,  as  may  be  seen  in  the  very  same  case 
in  the  34  prop.  b.  I,  it  was  thought  better  to  make  use  only  of 
the  4  ofb.  1. 

PROP.  XLV.     B.  I. 

The  straight  line  KM  is  proved  to  be  parallel  to  FL  from  the 
33  prop.;  whereas  KH  is  parallel  to  FG  by  construction,  and 
KHM,  FGL  have  been  demonstrated  to  be  straight  lines.  A 
corollary  is  added  from  Commandine,  as  being  often  used. 

PROP.  XIIL     B.  IL 

IN  this  proposition  only  acute  angled  triangles  are  mentioned,  Book  II. 
whereas  it  holds  true  of  every  triangle  :  and  the  demonstrations  s^^n'^^^ 
of  the  cases  omitted  are  added  ;  Commandine  and  Clavius  have 
likewise  given  their  demonstrations  of  these  cases. 

PROP.  XIV.     B.  II. 

In  the  demonstration  of  this,  some  Greek  editor  has  ignorant- 
ly  inserted  the  words  "  but  if  not,  one  of  the  two  BE,  ED  is  the 


S<i)S  NOTES. 

Book  II.  «  greater  ;  let  BE  be  the  greater,  and  pi'oduce  it  to  F,"  as  if  it- 
S.^'V^^  was  of  any  consequence  whether  the  greater  or  lesser  be  pror 

duced :  therefore,  instead  of  these  words,  there  ought  to  be  read 

only,  "  but  if  not,  produce  BE  to  F." 


PROP.  I.     B.  III. 

Book  III.  SEVERAL  authors, especially  among  the  modern  mathema- 
^-^"'>''^>*''  ticians  and  logicians,  inveigh  too  severely  against  indirect  or 
Apagogic  demonstrations,  and  sometimes  ignorantly  enough, 
not  being  aware  that  there  are  some  things  that  cannot  be  de- 
monstrated any  other  way  :  of  this  the  present  proposition  is  a 
very  clear  instance,  as  no  direct  demonstration  can  be  given  of 
it :  because,  besides  the  definition  of  a  circle,  there  is  no  princi- 
ple or  property  relating  to  a  circle  antecedent  to  this  problem, 
from  which  either  a  direct  or  indirect  demonstration  can  be  de- 
duced :  wherefore  it  is  necessary  that  the  point  found  by  the 
construction  of  the  problem  be  proved  to  be  the  centre  of  the 
circle,  by  the  help  of  this  definition,  and  some  of  the  preceding 
propositions  :  and  because,  in  the  demonstration,  this  proposi- 
tion must  be  brought  in,  viz.  straight  lines  from  the  centre  of  a 
circle  to  the  circumference  are  equal,  and  that  the  point  found 
by  the  construction  cannot  be  assumed  as  the  centre,  for  this  is 
the  thing  to  be  demonstrated  ;  it  is  manifest  some  other  point 
must  be  assumed  as  the  centre  ;  and  if  froni  this  assumption  an 
absurdity  follows,  as  Euclid  demonstrates  there  must,  then  it  is 
not  true  that  the  point  assumed  is  the  centre  ;  and  as  any  point 
whatever  was  assumed,  it  follows  that  no  point,  except  that  found 
by  the  construction,  can  be  the  centre,  from  which  the  necessity 
pf  an  indirect  demonstration  in  this  case  is  evident. 


PROP.  XIII.     B.  III. . 

As  it  is  much  easier  to  imagine  that  two  circles  may  touch 
one  another  within  in  more  points  than  one,  upon  the  same 
side,  than  upon  opposite  sides ;  the  figure  of  that  case  ought 
not  to  have  been  omitted;  but  the  construction  in  the  Greek 
text  would  not  have  suited  with  this  figure  so  well,  because  the 
centres  of  the  circles  must  have  been  placed  near  to  the  cir- 
cumferences :  on  which  account  another  construction  and  de- 
monstration is  given,  which  is  the  same  with  the  second  part 
pf   that  which  Campanus   h^s   translated   from    the  Arabic-, 


NOTES.  309 

where  without  any  reason  the  demonstration  is  divided  into  two  Book  III. 

PROP.  XV.     B.  III. 

The  co; J  verse  of  the  second  part  of  this  proposition  is  want- 
ing, though  in  the  preceding,  the  converse  is  added,  in  a  like 
case,  both  in  the  enunciation  and  demonstration  ;  and  it  is  now 
added  in  this.  Besides,  in  the  demonstration  of  the  first  part  of 
this  1 5th,  the  diameter  AD  (see  Commandine's  figure)  is  proved 
to  be  greater  than  the  straight  line  BC  by  means  of  another 
straight  line  MN  ;  whereas  it  may  be  better  done  without  it  : 
on  which  accounts  we  have  given  a  different  demonstration,  like 
to  that  which  Euclid  gives  m  the  preceding  14th,  and  to  that 
which  Theodosius  gives  in  prop.  6,  b.  l,of  his  Spherics,  in 
this  very  affair. 

PROP.  XVI.     B.  III. 

In  this  we  have  not  followed  the  Greek  nor  the  Latin  trans- 
lation literally,  but  have  given  what  is  plainly  the  meaning  of 
the  proposition,  without  mentioning  the  angle  of  the  semicircle, 
or  that  which  some  call  the  cornicular  angle  which  they  con- 
ceive to  be  made  by  the  circumference  and  the  straight  line 
which  is  at  right  angles  to  the  diameter,  at  its  extremity ;  which 
angles  have  furnished  matter  of  great  debate  between  some  of 
the  modern  geometers,  and  given  occasion  of  deducing  strange 
consequences  from  them,  whicn  are  quite  avoided  by  the  man- 
ner in  which  we  have  expressed  the  proposition.  And  in  like 
manner,  we  have  given  the  true  meaning  of  prop.  3 1,  b.  3,  with- 
out mentioning  the  angles  of  the  greater  or  lesser  segments  : 
these  passages,  Vieta,  with  good  reason,  suspects  to  be  adulte- 
rated, in  the  386th  page  of  his  Oper.  Math. 

PROP.  XX.     B.  III. 

The  first  words  of  the  second  part  of  this  demonstration, 
"  x.ix.XxcrB-u  3))  TtuXiv"  are  Avrong  translated  by  Mr.  Briggs  and 
Dr.  Gregory  "  Rursus  inclinetur ;"  for  the  translation  ought 
to  be  "  Rursus  inflectatur,"  as  Commandine  has  it :  a  straight 
line  is  said  to  be  inflected  either  to  a  straight  or  curve  line, 
when  a  straight  line  is  drawn  to  this  line  from  a  point,  and 
from  the  point  in  which  it  meets  it,  a  straight  line  making 
an  angle  with  the  former  is  drawn  to  another  point,  as  is  evi- 
dent from  the  90Lh  prop,  of  Euclid's  Data  :  for  this  the  whole 
line  betwixt  the  first  and  last  points,  is  inflected  or  broken  at 


310  NOTES._ 

Book  III.  the  point  of  inflection,  where  the  two  straight  lines  meet.  And 
^■^'■^''''"'W  in  the  like  sense  two  straight  lines  are  said  to  be  inflected  from 
two  points  to  a  third  point,  when  they  make  an  angle  at  this 
point ;  as  may  be  seen  in  the  description  given  by  Pappus  Alex- 
andrinus  of  Appollonius's  boohs  de  Locis  planis,  in  the  preface 
to  his  7th  book :  we  have  made  the  expression  fuller  from  the 
90th  prop,  of  the  Data. 


PROP.  XXI.     B.  III. 

There  are  two  cases  of  this  proposition,  the  second  of  which 
viz.  when  the  angles  are  in  a  segment  not  greater  than  a  semi- 
circle, is  wanting  in  the  Greek :  and  of  this  a  more  simple 
demonstration  is  given  than  that  which  is  in  Commandine,  as 
being  derived  only  from  the  first  case,  without  the  help  of  tri- 
angles. 

PROP.  XXIII.  and  XXIV.  B.  III. 

In  proposition  24  it  is  demonstrated,  that  the  segment  AFB 
must  coincide  with  the  segment  CFD,  (see  Commandine's 
figure),  and  that  it  cannot  fall  otherwise,  as  CGD,  so  as  to  cut 
the  other  circle  in  a  third  point  G,  from  this,  that,  if  it  did,  a 
circle  could  cut  another  in  more  points  than  two :  but  this 
ought  to  have  been  proved  to  be  impossible  in  the  23d  prop,  as 
well  as  that  one  of  the  segments  cannot  fall  within  the  other  : 
this  part  then  is  left  out  in  the  24th,  and  put  in  its  proper  place, 
the  23d  proposition. 

PROP.  XXV.     B.  III. 

This  proposition  is  divided  into  three  cases,  of  which  two 
have  the  same  construction  and  demonstration  ;  therefore  it  is 
now  divided  only  into  two  cases. 

PROP.  XXXIII.  B.  III. 

This  also  in  the  Greek  is  divided  into  three  cases,  of  which 
tv/o,  viz.  one  in  which  the  given  angle  is  acute,  and  the  other  in 
which  it  is  obtuse,  have  exactly  the  same  construction  and  de- 
7Bonstration  ;  on  which  account,  the  demonstration  of  the  last 
case  is  left  out  as  quite  superfluous,  and  the  addition  of  some 
unskilful  editor ;  besides  the  demonstration  of  the  case  when  the 
angle  given  is  a  right  angle,  is  done  a  roundabout  way,  and  is 
therefore  changed  to  a  more  simple  one,  as  was  done  by  Clavius. 


NOTES. 

PROP.  XXXV.    B.  III. 

As  the  25th  and  3. Id  propositions  are  divided  into  more  cases, 
so  this  35th  is  di\ided  into  fewer  cases  than  are  necessary.  Nor 
can  it  be  supposed  that  EucUd  omitted  them  because  they  are 
easy ;  as  he  has  given  the  case,  w^hich  by  far,  is  the  easiest  of 
them  all,  viz.  that  in  w^hich  both  the  straight  lines  pass  through 
the  centre :  and  in  the  following  proposition  he  separately  de- 
monstrates the  case  in  which  the  straight  line  passes  through 
the  centre,  and  that  in  which  it  does  not  pass  through  the  cen- 
tre :  so  that  it  seems  Theon,  or  some  other,  has  thought  them 
too  long  to  insert :  but  cases  that  require  different  demonstra- 
tions, should  not  be  left  out  in  the  Elements,  as  was  before  taken 
notice  of:  these  cases  are  in  the  translation  from  the  Arabic, 
and  are  now  put  into  the  text. 

PROP.  XXXVII.    B.  III. 

At  the  end  of  this,  the  words,  "  in  the  same  manner  it  may 
"  be  demonstrated,  if  the  centre  be  in  AC,"  are  left  out  as  the 
addition  of  some  ignorant  editor. 


DEFINITIONS  OF  BOOK  IV. 

WHEN  a  point  is  in  a  straight  line,  or  any  other  line,  this  Book  IV. 
point  is  by  the  Greek  geometers  said  xzs-TiirB-xiy  to  be  upon,  or  in  ^^'~^'">*^ 
that  line,  and  when  a  straight  line  or  circle  meets  a  circle  any 
way,  the  one  is  said  uTTTio-B-en  to  meet  the  other :  but  when  a 
strciight  line  or  circle  meets  a  circle  so  as  not  to  cut  it,  it  is  said 
{(pccTFTta-^-ui,  to  touch  the  circle  ;  and  these  two  terms  are  never 
promiscuously  used  by  them  :  therefore,  in  the  5th  definition 
of  book  4,  the  compound  upcKrTr.rxi  must  be  read,  instead  of  the 
simple  etTTTtiTcit:  and  in  the  1st,  2d,  3d,  and  6th  definitions  in 
Commandine's  translation,  "  tangit,"  must  be  read  instead  of 
"  contingit :"  and  m  the  2d  and  3d  definitions  of  book  3,  the 
same  change  must  be  made  :  but  in  the  Greek  text  of  proposi- 
tions nth,  12th,  13th,  18th,  19th,  book  3,  the  compound  verb 
is  to  be  put  for  the  simple. 

PROP.  IV.     B.  IV. 

In  this,  as  also  in  the  8th  and  1 3th  proposition  of  this  book, 
it  is  demonstrated  indirectly,  that  the  circle  touches  a  straight 
line  :  whereas  in  the  17th,  33d,  and  37th  propositions  of  book 
3,  the  same  thing  is  directly  demonstrated  :.  and  this  way  wc 


312  NOTES. 

Book  IV.  have  chosen  to  use  in  the  propositions  of  this  book,  as  it  is 
^•-^"^^^^w  shorter, 

PROP.  V.     B.  IV. 

The  demonstration  of  this  has  been  spoiled  by  some  unskil- 
ful hand  :  for  he  does  not  demonstrate,  as  is  necessary,  that  the 
two  straight  lines  which  bisect  the  sides  of  the  triangle  at  right 
angles  must  meet  one  another ;  and,  without  any  reason,  he  di- 
vides the  proposition  into  three  cases ;  whereas,  one  and  the 
same  construction  and  demonstration  serves  for  them  all,  as 
Campanus  has  observed  ;  which  useless  repetitions  are  now 
left  out :  the  Greek  text  also  in  the  corollary  is  manifestly  \'i- 
tiated,  where  mention  is  made  of  a  given  angle,  though  there 
neither  is,  nor  can  be  any  thing  in  the  proposition  relating  to  a 
given  angle. 

PROP.  XV.  and  XVI.     B.  IV. 

In  the  corollary  of  the  first  of  these,  the  words  equilateral 
and  equiangular  are  wanting  in  the  Greek  :  and  in  prop.  16, 
instead  of  the  circle  of  ABCD,  ought  to  be  read  the  circumfer- 
ence ABCD  :  where  mention  is  made  of  its  containing  fifteen 
equal  parts. 


DEF.  III.    B.  V. 

_  ,  y.  MANY  of  the  modern  mathematicians  reject  this  definition : 
^^.^.,'  the  very  learned  Dr.  Barrow  has  explained  it  at  large  at  the  end 
of  his  third  lecture  of  the  year  1666,  in  which  also  he  answers 
the  objections  made  against  it  as  well  as  the  subject  would  al- 
low :  and  at  the  end  gives  his  opinion  upon  the  whole,  as  fol- 
lows : 

"  I  shall  only  add,  that  the  author  had,  perhaps,  no  other 
"  design  in  making  this  definition,  than  (that  he  might  more 
"  fully  explain  and  embellish  his  subject)  to  give  a  general 
"  and  summary  idea  of  ratio  to  beginners,  by  premising 
•'  this  metaphysical  definition,  to  the  more  accurate  defini- 
"  tions  of  ratios  that  are  the  same  to  one  another,  or  one  of 
"  which  is  greater,  or  less  than  the  other :  I  call  it  a  meta- 
"  physical,  for  it  is  not  properly  a  mathematical  definition, 
"  since  nothing  in  mathematics  depends  on  it,  or  is  deduced, 
"  nor,  as  I  judge,  can  be  deduced  from  it :  and  the  defini- 
"  tion  of  analogy,  which  follows,  viz.  Analogy  is  the   simi- 


NOTES.  313 

**  litude  of  ratios,  is  of  the  same  kind,  and  can  serve  for  no  Book  V. 

"  purpose  in  mathematics,  but  only  to  give  beginners  some  ^-^"^^'^^i' 

"  general,  though  gross  and  confused  notion  of  analogy  :  but 

"  the  whole  of  the  doctrine  of  ratios,  and  the  whole  of  mathe- 

"  matics,  depend  upon  the  accurate  mathematical  definitions 

"  which  follow  this :  to  these  we  ought  principally  to  attend,  as 

"  the  doctrine  of  ratios  is  more  perfectly  explained  by  them ; 

"  this  third,  and  others  like  it,  may  be  entirely  spared  without 

"  any  loss  to  geometry  ;  as  Ave  see  in  the  7th  book  of  the  Ele- 

"  ments,  where  the  proportion  of  numbers  to  one  another  is 

"  defined,  and  treated  of,  yet  without  giving  any  definition  of 

"  the  ratio  of  numbers ;  though  such  a  definition  was  as  neces- 

"  sary  and  useful  to  be  given  in  that  book,  as  in  this  :  but  in- 

"  deed  shere  is  scarce  any  need  of  it  in  either  of  them:  though 

"  I  think  that  a  thing  of  so  general  and  abstracted  a  nature,  and 

"  thereby  the  more  difficult  to  be  conceived  and  explained,  can- 

"  not  be  more  commodiously  defined  than  as  the  author  has 

"  done :  upon  which  account  I  thought  fit  to  explain  it  at  large, 

"  and  defend  it  against  the  captious  objections  of  those  Avho 

"  attack  it."    To  this  citation  from  Dr.  Barrow  I  have  nothing 

to  add,  except  that  I  fully  believe  the  Sd  and  8th  definitions  are 

not  Euclid's,  but  added  by  some  unskilful  editor^ 


DEF.  XI.    B.  V. 

It  was  necessary  to  add  the  word  "  continual"  before  "  pro- 
"  portionals"  in  this  definition  ;  and  thus  it  is  cited  in  the  33d 
prop,  of  book  1 1 . 

After  this  definition  ought  to  have  followed  the  definition  of 
compound  ratio,  as  this  was  the  proper  place  for  it ;  duplicate 
and  triplicate  ratio  being  species  of  compound  ratio.  But  Theon 
has  made  it  the  5th  def.  of  book  6,  where  he  gives  an  absurd 
and  entirely  useless  definition  of  compound  ratio  :  for  this  rea- 
son we  have  placed  another  definition  of  it  betwixt  the  1 1th  and 
12th  of  this  book,  which,  no  doubt,  Euclid  gave  ;  for  he  cites  it 
expressly  in  prop.  23,  book  6,  and  which  Clavius,  Herigon,  and 
Barrow  have  likewise  given,  but  they  retain  also  Theon's, 
which  they  ought  to  have  left  out  of  the  Elements. 


DEF.  Xm.  B.  V. 

This,  and  the  rest  of  the  definitions  following,  contain  the  ex- 
plication of  some  terms  which  are  used  in  the  5th  and  following 
books;  which,  except  a  few,  are  easilv  enougli  understood  from. 

2   R      ■ 


iU  NOTES. 

Book  V.  the  propositions  of  this  book  where  they  are  first  mentioned  : 
^.^'-^^'^•^.^  they  seem  to  have  been  added  by  Theon,  or  some  other.  How- 
ever it  be,  they  are  explamed  something  more  distinctly  for  the 
sake  of  learners. 


PROP.  IV.     B.  V. 

In  the  construction  preceding  the  demonstration  of  this,  the 
words  d  £ry;^£,  any  whatever,  are  twice  wanting  in  the  Greek, 
as  also  in  the  Latin  translations  ;  and  are  now  added,  as  being 
wholly  necessary. 

Ibid,  in  the  demonstration  ;  in  the  Greek,  and  in  the  Latin 
translation  of  Commandine,  and  in  that  of  Mr.  Henry  Briggs, 
Avhich  was  published  at  London  in  1620,  together  with  the 
Greek  text  of  the  first  six  books,  which  translation  in  this  place 
is  followed  by  Dr.  Gregory  in  his  edition  of  Euclid,  there  is  this 
sentence  following,  viz.  "  and  of  A  and  C  have  been  taken 
"  equimultiples  K,  L  ;  and  of  B  and  D,  any  equimultiples 
"  whatever  («.'  irvx^)  M,  N  ;"  which  is  not  true,  the  words 
"  any  whatever,"  ought  to  be  left  out :  and  it  is  strange  that 
neither  Mr.  Briggs,  who  did  right  to  leave  out  these  words  in 
one  place  of  prop.  1 3  of  this  book,  nor  Dr.  Gregory  who  chang- 
ed them  into  the  word  "  some"  in  three  places,  and  left  them 
out  in  a  fourth  of  that  same  prop.  13,  did  not  also  leave  them 
out  in  this  place  of  prop.  4,  and  in  the  second  of  the  two  places 
where  they  occur  in  prop.  17,  of  this  book,  in  neither  of  which 
they  can  stand  consistent  with  truth  :  and  in  none  of  all  these 
])laces,  even  in  those  which  they  corrected  in  their  Latin  trans- 
lation, have  they  cancelled  the  words  »  £Tt;;^sin  the  Greek  text, 
as  they  ought  to  have  done. 

The  same  words  d  irvx?  are  found  in  four  places  of  prop.  1 1, 
of  this  book,  in  the  first  and  last  of  which  they  are  necessary, 
but  in  the  second  and  third,  though  they  are  true,  they  are  quite 
superfluous  ;  as  they  likewise  are  in  the  second  of  the  two  pla- 
ces in  which  til cy  are  found  in  the  12th  prop,  and  in  the  like 
places  of  prop.  22,  23,  of  this  book ;  but  are  wanting  in  the  last 
place  of  prop.  23,  as  also  in  prop.  25,  book  1 1. 


COR.  IV.  PROP.     B.  V. 

This  corollary  has  been  unskilfully  annexed  to  this  propo- 
sition, and  has  been  made  instead  of  the  legitimate  dem.on- 
stration,  which,  without  doubt,  Theon,  or  some  other  editor, 
has  taken  away,  not  from  this,  but  from  its  proper  place  in 


NOTES.  315 

this  book:  the  author  of  it  designed  to  demonstrate,  that  if  four  Book  V. 
magnitudes  E,  G,  F,  H  be  proportionals,  they  are  also  proper-  ^.-'■^'■>*-' 
tionals  inversely ;  that  is  G  is  to  E,  as  H  to  F  ;  which  is  true  ; 
but  the  demonstration  of  it  does  not  in  the  least  depend  upon 
this  4th  prop,  or  its  demonstration:  for,  when  he  says,  "  be- 
"  cause  it  is  demonstrated  that  if  K  be  greater  than  M,  L  is 
"  greater  than  N,"  £cc.  This  indeed  is  shown  in  the  demon- 
stration of  the  4th  prop,  but  not  from  this,  that  E,  G,  F,  H  are 
proportionals;  for  this  last  is  the  conclusion  of  the  proposition. 
Wherefore  these  words,  "  because  it  is  demonstrated,"  Sec.  are 
wholly  foreign  to  his  design :  and  he  should  have  proved,  that 
if  K  be  greater  than  M,  L  is  greater  than  N,  from  this,  that 
E,  G,  F,  H  are  proportionals,  and  from  the  5th  def.  of  this 
book  which  he  has  not ;  but  is  done  in  proposition  B,  which  we 
have  given  in  its  proper  place,  instead  of  this  corollary  ;  and 
another  corollary  is  placed  after  the  4th  prop,  which  is  often  of 
use  ;  and  is  necessary  to  the  demonstration  of  prop.  18  of  tliis 
book. 


PROP.  V.     B.  V. 


In  the  construction  which  precedes  the  demonstration  of  this 
proposition,  it  is  required  that  EB  may  be  the  same  multiple 
of  CG,  that  AE  is  of  CF  -.,  that  is,  that  EB  be  divided  into  as 
many  equal  parts,  as  there  are  parts  in  AE  equal  to  CF :  from 
which  it  is  evident,  that  this  construction  is  not  Euclid's  ;  for 
he  does  not  show  the  way  of  di\ading  straight  lines,  and  far  less 
other  magnitudes,  into  any  number  of  equal  parts,  until  the  9th 
pi'oposition  of  book  6 ;  and  he  never  requires  any  thing  to  be 
done  in  the  construction,  of  which  he  had  not  before  given  the 
method  of  doing.  For  this  reason,  we  have 
changed  the  construction  to  one,  which,  with- 
out doubt,  is  Euclid's,  in  which  nothing  is  re- 
quired but  to  add  a  magnitude  to  itself  a  certain  E— ' 
number  of  times ;  and  this  is  to  be  found  in  the 
translation  from  the  Arabic,  though  the  enuncia- 
tion of  the  proposition  and  the  demonstration 
are  there  very  much  spoiled.  Jacobus  Peletarius, 
who  was  the  first,  as  far  as  I  know,  who  took  no-  B  D 
tice  of  this  error,  gives  also  the  right  construc- 
tion in  his  edition  of  Euclid,  after  he  had  given  the  other  Avhich 
he  blames.  He  says,  he  Avould  not  leave  it  out,  because  it  Avas 
fine,  and  might  sharpen  one's  genius  to  invent  others  like  it ; 
whereas  there  is  not  the  least  difference  between  the  two  demon- 


C 


316  NOTES. 

Book  V.  strations,  except  a  single  word  in  the  construction,  which  very 
^^.^'^r^-m^  probably  has  been  owing  to  an  unskilful  librarian.  Clavius  like- 
wise gives  both  the  ways  ;  but  neither  he  nor  Peletarius  takes 
notice  of  the  reason  why  the  one  is  preferable  to  the  other. 

PROP.  VI.     B.  V. 

There  are  tAvo  cases  of  this  proposition,  of  which  only  the 
first  and  simplest  is  demonstrated  in  the  Greek  ;  and  it  is  pro- 
bable Theon  thought  it  was  sufficient  to  give  this  one,  since  he 
was  to  make  use  of  neither  of  them  in  his  mutilated  edition  of 
the  5th  book  ;  and  he  might  as  well  have  left  out  the  other,  as 
also  the  5th  proposition,  for  the  same  reason.  The  demonstra- 
tion of  the  other  case  is  now  added,  because  both  of  them,  as 
also  the  5th  proposition  are  necessary  to  the  demonstration  of 
the  18th  proposition  of  this  book.  The  translation  from  the 
Arabic  gives  both  cases  briefly. 


PROP.  A.     B.  V. 

This  proposition  is  fi-equently  used  by  geometers,  and  it  is 
necessary  in  the  25th  prop,  of  this  book,  31st  of  the  6th,  and 
34th  of  the  1 1th,  and  15th  of  the  12th  book.  It  seems  to  have 
been  taken  out  of  the  Elements  by  Theon,  because  it  appeared 
evident  enough  to  him,  and  others,  who  substitute  the  confused 
and  indistinct  idea  the  vulgar  have  of  proportionals,  in  place 
of  that  accurate  idea  which  is  to  be  got  from  the  5th  definition 
of  this  book.  Nor  can  there  be  any  doubt  that  Eudoxus  or  Eu- 
clid gave  it  a  place  in  the  Elements,  when  we  see  the  7th  and 
9th  of  the  same  book  demonstrated,  though  they  are  quite  as 
easy  and  evident  as  this.  Alphonsus  Borellus  takes  occasion 
from  this  proposition  to  censure  the  5th  definition  of  this  book 
very  severely,  but  most  unjustly.  In  p.  126  of  his  Euclid  restor- 
ed, printed  at  Pisa  in  1658,  he  says,  "  Nor  can  even  this  least 
t*  degree  of  knowledge  be  obtained  from  the  foresaid  property," 
viz.  that  which  is  contdned  in  5th  def.  5.  "  That,  if  four 
"  magnitudes  be  proportionals,  the  third  must  necessarily  be 
"  greater  than  the  fourth,  when  the  first  is  greater  than  the 
"  second ;  as  Clavius  acknowledges  in  the  1 6th  prop,  of  the 
"  5th  book  of  the  Elements."  But  though  Clavius  makes  no 
such  acknowledgment  expressly,  he  has  given  Eorellus  a  han- 
dle to  say  this  of  him  ;  because  when  Clavius,  in  the  above 
cited  place,  censures  Commandine,  and  that  very  justly,  for  de- 
monstrating this  proposition  by  help  of  the  16th  of  the  fifth  ; 
yet  he  himself  gives  no  demonstration  of  it,  but  thinks  it  plain 


NOTES. 


317 


from  the  nature  of  proportionals,  as  he  writes  in  the  end  of  the  Book  V. 
14th  and  1 6th  prop,  book  5  of  his  edition,  and  is  followed  by  He-  ^--'"^•''^^m/ 
rigon  in  Schol.  1,  prop.  14th,  book  3,  as  if  there  was  any  nature 
of  proportionals  antecedent  to  that  which  is  to  be  derived  and 
understood  from  the  definition  of  them.  And,  indeed,  though  it 
is  very  easy  to  give  a  right  demonstration  of  it,  nobody,  as  far  as 
I  know,  has  given  one,  except  the  learned  Dr.  Barrow,  who,  in 
answer  to  Borellus's  objection,  demonstrates  it  indirectly,  but 
very  briefly  and  clearly,  from  the  5th  definition,  in  the  322d 
page  of  his  Lect.  Mathem.  from  which  definition  it  may  also  be 
easily  demonstrated  directly.  On  which  account  we  have  placed 
it  next  to  the  propositions  concerning  equimultiples. 


PROP.  B.     B.  V. 

This  also  is  easily  deduced  from  the  5th  def.  b,  5,  and  therc- 
fore  is  placed  next  to  the  other ;  for  it  was  very  ignorantly  made 
a  corollary  from  the  4th  prop,  of  this  book.  See  the  note  on 
that  corollary. 


PROP.  C.     B.  V. 

This  is  frequently  made  use  of  by  geometers,  and  is  necessary 
to  the  5th  and  6th  propositions  of  the  10th  book.  Clavius,  in 
his  notes  subjoined  to  the  8th  def.  of  book  5,  demonstrates  it 
only  in  numbers,  by  help  of  some  of  the  propositions  of  the  7th 
book  ;  in  order  to  demonstrate  the  property  contained  in  the  5th 
definition  of  the  5th  book,  when  applied  to  numbers,  from  the 
property  of  proportionals  contained  in  the  20th  def.  of  the  7th 
book.  And  most  of  the  commentators  judge  it  difficult  to  prove 
that  four  magnitudes  which  are  proportionals  according  to  the 
20th  def.  of  7th  book,  are  also  proportionals  according  to  the 
5tli  def.  of  5th  book.     But  this  is  easily  made  out,  as  follows. 

First,  If  A,  B,  C,  D  be  four 
magnitudes,  such  that  A  is  the 
same  multiple,  or  the  same  part 
of  B,  which  C  is  of  D  ;  A,  B, 
C,  D  are  proportionals.  This  is 
demonstrated  in  proposition  C. 

Secondly,  If  AB  contain  the 
same  parts  of  CD,  that  EF  does 
of  GH :  in  this  case  likewise  AB 
is  to  CD,  as  EF  to  GH. 


D 


K-V 


H 


Lf 


\     C     E    G     M 


318 


NOTES. 


B 


H 


D 


Kt 


A      C    E    G     M 


Book  V.       Let  CK  be  a  part  of  CD,  and  GL  the  same  part  of  GH  ; 

v^'v"'^^  and  let  AB  be  the  same  multiple  of 
CK,  that  EF  is  of  GL  :  therefore, 
by  prop.  C,  of  5th  book,  AB  is  to 
CK,  as  EF  to  GL:  and  CD,  GH 
are  equimultiples  of  CK,  GL  the 
second  and  fourth  ;  wherefore,  by 
cor.  prop.  4,  book  5,  AB  is  to  CD, 
as  EF  to  GFL 

And  if  four  magnitudes  be  pro- 
portionals according  to  the  5th  def. 

of  book  5,  they  are  also  proportionals  according  to  the  20th  def. 
of  book  7. 

First,  If  A  be  to  B,  as  C  to  D  ;  then  if  A  be  any  multiple  or 
part  of  B,  C  is  the  same  multiple  or  part  of  D,  by  prop.  D,  of 
book  5. 

Next,  If  AB  be  to  CD,  as  EF  to  GH  ;  then  if  AB  contains 
any  parts  of  CD,  EF  contains  the  same  parts  of  GH  :  for  let 
CK  be  a  part  of  CD,  and  GL  the  same  part  of  GH,  and  let 
AB  be  a  multiple  of  CK  ;  EF  is  the  same  multiple  of  GL  : 
take  M  the  same  multiple  of  GL  that  AB  is  of  CK  ;  there- 
fore by  prop.  C,  of  book  5,  AB  is  to  CK,  as  M  to  GL  ;  and  CD, 
GH  are  equimultiples  of  CK,  GL  :  wherefore  by  cor.  prop.  4, 
b.  5,  AB  is  to  CD,  as  M  to  GH.  And,  by  the  hypothesis,  AB 
is  to  CD,  as  EF  to  GH  ;  therefore  M  is  equal  to  EF,  by  prop. 
9,  book  5,  and  consequently  EF  is  the  same  multiple  of  GL  that 
ABisofCK. 


PROP.  D.  B.  V. 

This  is  not  unfrequently  used  in  the  demonstration  of  other 
propositions,  and  is  necessary  in  that  of  prop.  9,b.  6.  It  seems 
Theon  has  left  it  out  for  the  reason  mentioned  in  the  notes  at 
prop. A. 

PROP.  VIII.     B.  V. 


In  the  demonstration  of  this,  as  it  is  now  in  the  Greek, 
there  are  two  cases  (see  the  demonstration  in  Hervagius,  or 
Dr.  Gregory's  edition),  of  which  the  first  is  that  in  which  AE 
is  less  than  EB ;  and  in  this  it  necessarily  follows,  that  H© 
the  multiple  of  EB  is  greater  than  ZH,  the  same  multiple  of 
AE,  which  last  multiple,  by  the  construction  is  greater  than  A  : 
whence  also  H©  must  be  greater  than  A.  But  in  the  second 
case,  viz.  that  in  which  EB  is  less  than  AE,  though  ZH  be 


NOTES. 


;i9 


greater  than  A,  yet  H0  may  be  less  than  the  same  A ;  so  that  Book  V. 
there  cannot  be  taken  a  multiple  of  A  which  is  the  first  that  is  v.^~v<^^*. 
greater  than  K  or  H0,  because  A  itself  is  greater  than  it :  up- 
on this  account,  the  author  of  this  demonstration  found  it  ne- 
cessary to  change  one  part  of  the  construction  that  was  made 
use  of  in  the  first  case :  but  he  has,  without  any  necessity, 
changed  also  another  part  of  it,  viz.  when  he  orders  to  take  N 
that  multiple  of  A  which  is  the       y  '. 

first  that  is  greater  than  ZH  ;  , 

for  he  might  have  taken  that 
multiple  of  A  which  is  the  first  . 

that  is  gi-eater  than  Ho,  or  K,   H— 
as  was  done  in  the  first  case  : 

he  likewise  brings  in  this  K  E  — .  H-^ 

into  the  demonstration  of  both 

cases,  without  any  reason  ;  for  E  4- 

it  serves  to  no  purpose  but  to 
lengthen    the     demonstration. 

There  is  also    a    third  case,       ©        B    A  0       B    A 

which  is  not  mentioned  in  this  demonstration,  viz.  that  in  which 
AE  in  the  first,  or  EB  in  the  second  of  the  two  other  cases,  is 
greater  than  D ;  and  in  this  any  equimultiples,  as  the  doubles, 
of  AE,  KB  are  to  be  taken,  as  is  done  in  this  edition,  where  all 
the  cases  are  at  once  demonstrated  :  and  from  this  it  is  plain 
that  Theon,  or  some  other  unskilful  editor,  has  vitiated  this 
proposition. 


PROP.  IX.    B.  V. 

Of  this  there  is  given  a  more  explicit  demonstration  than 
that  which  is  now  in  the  Elements. 


PROP.  X.     B.  V. 

It  was  necessary  to  give  another  demonstration  of  this  pro- 
position, because  that  which  is  in  the  Greek  and  Latin,  or  other 
editions,  is  not  legitimate  :  for  the  words  greater,  the  same,  or 
equal,  lesser,  have  a  quite  different  meaning  when  applied  to 
magnitudes  and  ratios,  as  is  plain  from  the  5  th  and  7  th  defini- 
tions of  book  5.  By  the  help  of  these  let  us  examine  the  de- 
monstration of  the  10th  prop,  which  proceeds  thus  :  "  Let  A 
*'  have  to  C  a  greater  ratio  than  B  to  C ;  I  say  that  A  is  greater 
'<  than  B.  For  if  it  is  not  greater,  it  is  either  equal,  or  less. 
"  But  A  cannot  be  equal  to  B,  because  then  each  of  them 
"  would  have  the  same  ratio  to  C  ;  but  they  have  not.  There- 
"  fore  A  is  not  equal  to  B."     The  force  of  v/hich  reasoning  is 


32Q  NOTES. 

Book  V.  this,  if  A  had  to  C,  the  same  ratio  that  B  has  to  C,  then  if 
*«^''^"'''^*^  any  equimultiples  whatever  of  A  and  B  be  taken,  and  any- 
multiple  whatever  of  C  ;  if  the  multiple  of  A  be  greater  than 
the  multiple  of  C,  then,  by  the  5th  def.  of  book  5,  the  multiple 
of  B  is  also  greater  than  that  of  C ;  but,  from  the  hypothesis 
that  A  has  a  greater  ratio  to  C,  than  B  has  to  C,  there  must, 
by  the  7th  def.  of  book  5,  be  certain  equimultiples  of  A  and  B, 
and  some  multiple  of  C  such,  that  the  multiple  of  A  is  greater 
than  the  multiple  of  C,  but  the  multiple  of  B  is  not  greater 
than  the  same  multiple  of  C  ;  and  this  proposition  directly 
contradicts  the  preceding  ;  wherefore  A  is  not  equal  to  B. 
The  demonstration  of  the  10th  prop,  goes  on  thus  :  "  But  nei- 
"  ther  is  A  less  than  B  ;  because  then  A  would  have  a  less  ra- 
"  tio  to  C,  than  B  has  to  it :  but  it  has  not  a  less  ratio,  there- 
"  fore  A  is  not  less  than  B,"  &c.  Here  it  is  said,  that "  A 
"  would  have  a  less  ratio  to  C,  than  B  has  to  C,"  or,  which 
is  the  same  thing,  that  B  would  have  a  greater  ratio  to  C, 
than  A  to  C;  that  is,  by  7th  def.  book  5,  there  mvist  be  some 
equimultiples  of  B  and  A,  and  some  multiple  of  C  such,  that 
the  multiple  of  B  is  greater  than  the  multiple  of  C,  but  the 
multiple  of  A  is  not  greater  than  it :  and  it  ought  to  have 
been  proved  that  this  can  never  happen  if  the  ratio  of  A  to 
C  be  greater  than  the  ratio  of  B  to  C  ;  that  is,  it  should  have 
been  proved,  that,  in  this  case,  the  multiple  of  A  is  always 
greater  than  the  multiple  of  C,  whenever  the  multiple  of  B  is 
greater  than  the  multiple  of  C  ;  for,  when  this  is  demonstrated, 
it  will  be  evident  that  B  cannot  have  a  greater  ratio  to  C,  than 
A  has  to  C,  or,  which  is  the  same  thing,  that  A  cannot  have  a 
less  ratio  to  C,  than  B  has  to  C  :  but  this  is  not  all  proved  in 
the  10th  proposition  ;  but  if  the  10th  were  once  demonstrated, 
it  would  immediately  follow  from  it,  but  cannot  without  it  be 
easily  demonstrated,  as  he  that  tries  to  do  it  will  find.  Where- 
fore the  10th  proposition  is  not  sufliciently  demonstrated.  And 
it  seems  that  he  who  has  given  the  demonstration  of  the  10th 
proposition  as  we  now  have  it,  instead  of  that  which  Eudoxus 
or  Euclid  had  given,  has  been  deceived  in  applying  what  is 
manifest,  when  understood  of  magnitudes,  unto  ratios,  viz.  that 
a  magnitude  cannot  be  both  greater  and  less  than  another. 
That  those  things  which  are  equal  to  the  same  arc  equal  to 
one  another,  is  a  most  evident  axiom  when  vmderstood  of 
magnitudes  ;  yet  Euclid  does  not  make  use  of  it  to  infer  that 
those  ratios  which  are  the  same  to  the  same  i-atio,  are  the  same 
to  one  another ;  but  explicitly  demonstrates  this  in  prop.  11, 
of  hook  5 .  The  demonstration  we  have  given  of  the  1 0th  prop,  is 


NOTES. 


321 


lio  doubt  the  same  with  that  of  Eudoxus  or  Euclid,  as  it  is  im-  Book  V. 
mediately  and  directly  derived  from  the  definition  of  a  greater  v-^'v''^>«-' 
Tatio,  viz.  the  7.  of  the  5. 

The  abovementioned  proposition,  viz.  If  A  have  to  C  a 
greater  ratio  than  B  to  C  ;  and  if  of  A  and 
B  there  be  taken  certain  equimultiples,  and 
some  multiple  of  C ;  then  if  the  multiple 
of  B  be  greater  than  the  multiple  of  C,  the 
multiple  of  A  is  also  greater  than  the 
same,  is  thus  demonstrated. 

Let  D,  E  be  equimultiples  of  A,  B,  and 
F  a  multiple  of  C,  such,  that  E  the  multiple 
of  B  is  greater  than  F ;  D  the  multiple  of 
A  is  also  greater  than  F. 

Because  A  has  a  greater  ratio  to  C,  than 
B  to  C,  A  is  greater  than  B,  by  the  10th 
prop.  B.  5;  therefore  D  the  multiple  of 
A  is  greater  than  E  the  same  multiple  of 
B :  and  E  is  greater  than  F ;  much  more 
therefoi'e  D  is  greater  than  F. 


A      C       B      C 


D 


PROP.  XIII.     B.  V. 

In  Commandine's,Briggs's,  and  Gregory's  translations,  at  the 
beginning  of  this  deinonstration,  it  is  said,  "  And  the  multi- 
"  pie  of  C  is  greater  than  the  multiple  of  D  ;  but  the  multiple 
"  of  E  is  not  greater  than  the  multiple  of  F ;"  which  words 
are  a  literal  translation  from  the  Greek :  but  the  sense  evidently 
requires  that  it  be  read,  "  so  that  the  multiple  of  C  be  greater 
"  than  the  multiple  of  D  ;  but  the  multiple  of  E  be  not  greater 
"  than  the  multiple  of  F."  And  thus  this  place  v,  as  restored  to 
the  true  i^eading  in  the  first  editions  of  Commandine's  Euclid, 
printed  in  8  vo.  at  Oxford;  but  in  the  later  editions,  at  least  in 
that  of  1747,  the  error  of  the  Greek  text  Avas  kept  in. 

There  is  a  corollary  added  to  prop.  13,  as  it  is  necessary  to 
the  20th  and  2 1st  prop,  of  this  book,  and  is  as  useful  as  the 
proposition. 


PROP.  XIV.     B.  V 


The  two  cases  of  this,  which  are  not  in  the  Greek,  are  add- 
ed; the  demonstration  of  them  not  being  exactly  the  same  with 
that  of  the  first  case. 

2  S 


NOTES. 


PROP.  XVII.     B.  V. 


The  order  of  the  words  in  a  clause  of  this  is  changed  to  one 
more  natural:  as  was  also  done  in  prop.  1. 


PROP.  XVIII.     B.  V. 

The  demonstration  of  this  is  none  of  Euclid's,  nor  is  it  legi- 
timate ;  for  it  depends  upon  this  hypothesis,  that  to  any  three 
magnitudes,  two  of  which,  at  least,  are  of  the  same  kind, 
there  may  be  a  fourth  proportional ;  which,  if  not  proved,  the 
demonstration  now  in  the  text  is  of  no  force :  but  this  is  as- 
sumed without  any  proof;  nor  can  it,  as  far  as  I  am  able  to 
discern,  be  demonstrated  by  the  propositions  preceding  this ; 
so  far  is  it  from  deserving  to  be  reckoned  an  axiom,  as  Cla- 
vius,  after  other  commentators,  would  have  it,  at  the  end  of 
the  definitions  of  tlie  5th  book.  Euclid  does  not  demonstrate 
it,  nor  does  he  show  how  to  find  the  fourth  propoitional,  be- 
fore the  12th  prop,  of  the  6th  book:  and  he  never  assumes  any 
thing  in  the  demonstration  of  a  proposition,  which  he  had  not 
before  demonstrated.;  at  least,  he  assumes  nothing  the  existence 
of  which  is  not  evidently  possible ;  for  a  certain  conclusion  can 
never  be  deduced  by  the  means  of  an  uncertain  proposition : 
vipon  this  account,  we  have  given  a  legitimate  demonstration 
of  this  proposition  instead  of  that  in  the  Greek  and  other  edi- 
tions, which  very  probably  Theon,  at  least  some  other,  has 
put  in  the  place  of  Euclid's,  because  he  thought  it  too  prolix : 
and  as  the  17th  prop,  of  v/hich  this  18th  is  the  converse,  is  de- 
monstrated by  help  of  the  1st  and  2d  propositions  of  this  book; 
so,  in  the  demonstration  now  given  of  the  18th,  the  5th  prop, 
and  both  cases  of  the  6th  are  necessary,  and  these  two  propo- 
sitions are  the  converses  of  the  1st  and  2d.  Now  the  5th  and 
6th  do  not  enter  into  the  demonstration  of  any  proposition  in 
this  book  as  we  now  have  it ;  nor  can  they  be  of  use  in  any 
proposition  of  the  Elements,  except  in  this  18th,  and  this  is  a 
manifest  proof,  that  Euclid  made  use  of  them  in  his  demon- 
stration of  it,  and  that  the  demonstration  now  given,  which  is 
exactly  the  converse  of  that  of  the  1 7th,  as  it  ought  to  be,  dif- 
fers nothing  from  that  of  Eudoxus  or  Euclid :  for  the  5th  and 
6th  have  undoubtedly  been  put  into  the  5th  book  for  the  sake 
of  some  propositions  in  it,  as  all  the  other  propositions  about 
eq\iimultiples  have  been. 

Hieronymus  Saccherius,  in  his  book  named  Euclides  ab  om- 
ni  nsivo  viiidicutus,  printed  at.  Milan,  anno   1733,  in  4to,  ac- 


NOTES.  32: 

knowledges  this  blemish  in  the  demonstration  of  the  18th,  and  Book  V. 
that  he  may  remove  it,  and  render  the  demonstration  we  now  ^•^'■^''^>*. 
have  of  it  legitimate,  he  endeavours  to  demonstrate  the  follow- 
ing proposition,  which  is  in  page  1 1 5  of  his  book,  viz. 

"  Let  A,  B,  C,  D  be  four  magnitudes,  of  which  the  two 
"  first  are  of  one  kind,  and  also  the  two  others  either  of  the 
"  same  kind  with  the  two  first,  or  of  some  other  the  same 
"  kind  with  one  another.  I  say  the  ratio  of  the  third  C  to  the 
"  fourth  D,  is  either  equal  to,  or  greater,  or  less  than  the  I'atio 
«  of  the  first  A  to  the  second  B." 

And  after  two  propositions  premised  as  lemmas,  he  proceeds 
thus. 

"  Either  among  all  the  possible  equimultiples  of  the  first 
"  A,  and  of  the  third  C,  and,  at  the  same  time,  among  all 
"  the  possible  equimultiples  of  the  second  B,  and  of  the 
"  fourth  D,  there  can  be  found  some  one  multiple  EF  of  the 
"  first  A,  and  one  IK  of  the  second  B,  that  are  equal  to  one 
"  another;  and  also,  in  the  same  case,  some  one  multiple 
"  GH  of  the  third  C  equal  to  LM  the  multiple  of  the  fourth 
"  D,  or  such  equality  is  no  where  to  be  found.  If  the  first 
"  case    happen, 

"  [i.  e.  if  such     A E F 

"  equality  is  to 

"  be  found]  it  is     B I K 

"  manifest  from 

«  what    is    be-     C G H 

"  fore    demon- 

"  strated,     that     D L M 

"  A  is  to  B  as 

"  C  to  D;  but  if  such  simultaneous  equality  be  not  to  be 
"  found  upon  both  sides,  it  will  be  found  either  upon  one 
"  side,  as  upon  the  side  of  A  [and  B ;]  or  it  will  be  found 
"  upon  neither  side  ;  if  the  first  happen ;  therefore,  from 
"  Euclid's  definition  of  greater  and  lesser  ratio  foregoing, 
"  A  has  to  B,  a  greater  or  less  ratio  than  C  to  D ;  accord- 
"  ing  as  GH  the  multiple  of  the  third  C  is  less,  or  greater 
"  than  LM  the  multiple  of  the  fourth  D  :  but  if  the  second 
"  case  happen  ;  therefore  upon  the  one  side,  as  upon  the  side 
"  of  A  the  first  and  B  the  second,  it  may  happen  that  the 
"  multiple  EF,  [viz.  of  the  first]  may  be  less  than  IK  the 
"  multiple  of  the  second,  while,  on  the  contrary,  upon  the  other 
"  side,  [viz.  of  C  and  D]  the  multiple  GH  [of  the  third  C]  is 
"  greater  than  the  other  multiple  LM  [of  the  foui-th  D  :]  and 
"  then,  from  the  same  definition  of  Euclid,  the  ratio  of  the  first 
"  A  to  the  second  B,  is  less  than  the  ratio  of  the  third  C  to  the 
*'  fourth  D ;  or  on  the  contrary. 


J24  ^OTES. 

Book  v.  "  Therefore  the  axiom  [i.  e.  the  proposition  before  set  down], 
V*^^^^"""^-^  "  remains  demonstrated,"  8cc. 

Not  in  the  least ;  but  it  remains  still  undemonstrated :  for 
what  he  says  may  happen,  may,  in  innumerable  cases,  never 
happen  ;  and  therefore  his  demonstration  does  not  hold :  for 
example,  if  A  be  the  side,  and  B  the  diameter  of  a  square  ; 
and  C  the  side,  and  D  the  diameter  of  another  square  ;  there 
•  can  in  no  case  be  any  multiple  of  A  equal  to  any  of  B  ;  nor 
any  one  of  C  equal  to  one  of  D,  as  is  well  known  ;  and 
yet  it  can  never  happen  that  when  any  multiple  of  A  is  greater 
than  a  multiple  of  B,  the  multiple  of  C  can  be  less  than  the  mul- 
tiple of  D,  nor  when  the  multiple  of  A  is  less  than  that  of  B, 
the  multiple  of  C  can  be  greater  than  that  of  D,  viz.  taking 
equimultiples  of  A  and  C,  and  equimultiples  of  B  and  D  :  for 
A,  B,  C,  D  are  proportionals  ;  and  so  if  the  multiple  of  A  be 
greater.  Sec.  than  that  of  B,  so  must  that  of  C  be  greater.  Sec. 
than  that  of  D ;  by  5th  def.  b.  5. 

The  same  objection  holds  good  against  the  demonstration 
which  some  give  of  the  1st  prop,  of  the  6th  book,  v.'hich  we 
have  made  against  this  of  the  18th  prop,  because  it  depends 
upon  the  banie  insufficient  foundation  with  the  other, 


PROP.  XIX.     B.  V. 

A  corollary  is  added  to  this,  which  is  as  frequently  used  as 
the  proposition  itself.  The  covoUary  which  is  subjoined  to  it 
in  the  Greek,  plainly  shows  that  the  5th  book  has  been  vitiated 
by  editors  who  were  not  geometers :  for  the  conversion  of 
ratios  does  not  depend  upon  this  19th,  and  the  demonstration 
which  several  of  the  commentators  on  Euclid  gave  of  conver- 
sion, is  not  legitimate,  as  Clavius  has  rightly  observed,  who 
has  given  a  good  demonstration  of  it  which  Ave  have  put  in  pro- 
position E ;  but  he  makes  it  a  corollary  from  the  19th,  and  be- 
fpns  it  Avith  the  words,  "  Hence  it  easily  follows,"  though  it  does 
not  at  all  follow  from  it. 


PROP.  XX.  XXI.  XXII.  XXIII.  XXIV.     B.  V. 

The  demonstrations  of  the  20th  and  21st  propositions,  are 
shorter  than  those  Euclid  gives  of  easier  propositions,  either 
in  the  preceding,  or  following  books :  wherefore  it  Avas  pro- 
per to  make  them  more  explicit,  and  the  22d  and  23d  propo-> 
hitions  are,  as  they  ought  to  be,  extended  to  any  number  of 


NOTES.  325 

Hiagnitudes :  and,  in  like  manner,  may  the  24th  be,  as  is  taken   Book  V. 
notice  of  in  a  corollary  ;  and  another  corollary  is  added,  as  use-  v^^n^^*^ 
fill  as  the  proposition,  and  the  words  "  any  whatever"  are  sup- 
plied near  the  end  of  prop.  23,  which  are  wanting  in  the  Greek 
text,  and  the  translations  from  it. 

In  a  paper  writ  by  Phillippus  Naudxus,  and  published  after 
his  death,  in  the  History  of  the  Royal  Academy  of  Sciences  of 
Berlin,  anno  1745,  page  50,  the  23d  prop,  of  the  5th  book  is 
censured  as  being  obscurely  enunciated,  and,  because  of  this, 
prolixly  demonstrated:  the  enunciation  there  given  is  not  Eu- 
clid's, but  Tacquet's,  as  he  acknowledges,  which,  though  not  so 
well  expressed,  is,  upon  the  matter,  the  same  with  that  which  is 
now  in  the  Elements.  Nor  is  there  any  thing  obscure  in  it, 
though  the  author  of  the  paper  has  set  clown  the  proportionals 
in  a  disadvantageous  order,  by  which  it  appears  to  be  obscure: 
but,  no  doubt,  Euclid  enunciated  this  23d,  as  well  as  the  2 2d, 
so  as  to  extend  it  to  any  number  of  magnitudes,  which  taken 
two  and  two  are  proportionals,  and  not  of  six  only ;  and  to  this 
general  case  the  enunciation  which  Naudxus  gives,  cannot  be 
well  applied. 

The  demon;.tration  which  is  given  of  this  23d,  in  that  paper, 
is  quite  wrong;  because,  if  the  proportional  magnitudes  be 
plane  or  solid  figures,  there  can  no  rectangle  (which  he  impro- 
perly calls  difiroduct)  be  conceived  to  be  made  by  any  two  of 
them :  and  if  it  should  be  said,  that  in  this  case  straight  lines  are 
to  be  taken  which  are  proportional  to  the  figures,  the  demon- 
stration would  this  way  become  much  longer  than  Euclid's :  but, 
even  though  his  demonstration  had  been  right,  who  does  not 
see  that  it  could  not  be  made  use  of  in  the  5th  book. 


PROP.  F,  G,  H,  K.     B.  V. 

These  propositions  are  annexed  to  the  5th  book,  because  they 
are  frequently  made  use  of  by  both  ancient  and  modern  geome- 
ters :  and  in  many  cases  compound  ratios  cannot  be  brought 
into  demonstration,  without  making  use  of  them. 

Whoever  desires  to  see  the  doctrine  of  ratios  delivered  in  this 
5th  book  solidly  defended,  and  the  arguments  brought  against 
it  by  And.  Tacquet,  Alph.  Borellus,  and  others,  fully  refuted, 
may  read  Dr.  Barrow's  Mathematical  Lectures,  viz.  the  7th  and 
8th  of  the  year  1666. 

The  5th  book  being  thus  corrected,  I  most  readily  agree  to 
what  the  learned  Dr.  Barrow  says,*  "  That  there  is  nothing 
"  in  the  whole  body  of  the  Elements  of  a  more  subtile  invention, 
*  Pag-e  336. 


326  NOTES. 

Book  V,  «  nothing  more  solidly  established,  and  more  accurately  hand- 

^-^'^'"^ii'  "  led,  than  the  doctrine  of  proportionals."     And  there  is  some 

ground  to  hope,  that  geometers  will  think  that  this  could  not 

have  been  said  with  as  good  reason,  since  Theon's  time  till  the 

present. 


DEF.  II.  AND  V.  OF  B.  VI. 

Book  VI.  THE  2d  definition  does  not  seem  to  be  Euclid's,  but  some  \in 
^•^~'^'^>»-^  skilful  editor's :  for  there  is  no  mention  made  by  Euclid,  nor,  as 
far  as  I  know,  by  any  other  geometer,  of  reciprocal  figures:  it 
is  obscurely  expressed,  which  made  it  proper  to  render  it  more 
distinct :  it  would  be  better  to  put  the  following  definition  in 
place  of  it,  viz. 


DEF.  II. 

Two  magnitudes  are  said  to  be  reciprocally  proportional  to 
two  others,  when  one  of  the  first  is  to  one  of  the  other  magni- 
tudes, as  the  remaining  one  of  the  last  two  is  to  the  remaining 
one  of  the  first. 

But  the  5th  definition,  which  since  Theon's  time,  has  been 
kept  in  the  Elements,  to  the  great  detriment  of  learners,  is  now 
justly  thrown  out  of  them,  for  the  reason  given  in  the  notes  on 
the  23d  prop,  of  this  book. 


PROP.  I.  and  II.     B.  VI. 

To  the  first  of  these  a  corollary  is  added,  which  is  often  used : 
and  the  enunciation  of  the  second  is  made  more  general. 


PROP.  III.  B.  VI. 

A  second  case  of  this,  as  useful  as  the  first,  is  given  in  prop. 
A:  viz.  the  case  in  which  the  exterior  angle  of  a  triangle  is  bi- 
sected by  a  straight  line :  the  demonstration  of  it  is  very  like  to 
that  of  the  first  case,  and  upon  this  account  may,  probably,  have 
been  left  out,  as  also  the  enunciation,  by  some  unskilful  editor : 
at  least,  it  is  certain,  that  Pappus  makes  use  of  this  case,  as  an 
elementary  proposition,  without  a  demonstration  of  it,  m  prop. 
39  of  his  7th  book  of  Mathematical  Collection's. 


NOTES. 


PROP.  VII.     B.  VI. 

To  this  a  case  is  added  which  occurs  not  unfrequently  in  de- 
monstrations. 

PROP.  VIII.     B.   VI. 

It  seems  plain  that  some  editor  has  changed  the  demonstra- 
tion that  Euclid  gave  of  this  proposition  :  for,  after  he  has  de- 
monstrated, that  the  triangles  are  equitingular  to  one  another, 
he  particularly  shows  that  their  sides  about  the  equal  angles  are 
proportionals,  as  if  this  had  not  been  done  in  the  demonstration 
of  the  4th  prop,  of  this  book  :  this  superfluous  part  is  not  found 
in  the  translation  from  the  Arabic,  and  is  now  left  out. 


PROP.  IX.    B.  VI. 

This  is  demonstrated  in  a  particular  case,  viz.  that  in  which 
the  thii'd  part  of  a  straight  line  is  required  to  be  cut  off ;  which 
is  not  at  all  like  Euclid's  manner :  besides,  the  author  of  the 
demonstration,  from  four  magnitudes  being  proportionals,  con- 
cludes that  the  'bird  of  them  is  the  same  multiple  of  the  fourth, 
which  the  first  is  of  the  second  ;  now,  this  is  no  where  demon- 
strated in  the  5th  book,  as  we  now  have  it :  but  the  editor  as- 
sumes it  from  the  confused  notion  which  the  vulgar  have  of  pro- 
portionals :  on  this  account,  it  was  necessary  to  give  a  general 
and  legitimate  demonstration  of  this  proposition. 


PROP.  XVIII.     B.  VI. 

The  demonstration  of  this  seems  to  be  vitiated:  for  tht- 
proposition  is  demonstrated  only  in  the  case  of  quadrilateral 
figures,  without  mentioning  how  it  may  be  extended  to  figures 
of  five  or  more  sides  :  besides,  from  two  triangles  being  equi- 
angular, it  is  inferred,  that  a  side  of  the  one  is  to  the  homolo- 
gous side  of  the  other,  as  another  side  of  the  first  is  to  the 
side  homologous  to  it  of  the  other,  without  permutation  of  the 
proportionals ;  which  is  contrary  to  Euclid's  manner,  as  is 
clear  from  the  next  proposition  :  and  the  sam^e  fault  occurs 
again  in  the  conclusion,  where  the  sides  about  the  equal  angles 
are  not  shown  to  be  proportionals,  by  reason  of  again  neglect- 
ing permutation.  On  these  accounts,  a  demonstration  is  given 
in  Euclid's  manner,  like  to  that  he  makes  use  of  in  the  "Otl> 


328  NOTES. 

Book  VI.   prop,  of  this  book  ;  and  it  is  extended  to  five-sided  figures,  by 
^-^'^•^'^'^m^  whicij  it  may  be  seen  how  to  extend  it  to  figures  of  any  number 
of  sides. 


PROP.  XXIII.     B.  VI. 

Nothing  is  usually  reckoned  more  difficult  in  the  elements 
of  geometry  by  learners,  than  the  doctrine  of  compound  ra- 
tio, which  Thcon  has  rendered  absurd  and  ungcometrical,  by 
substituting  the  5th  definition  of  the  6th  book  in  place  of  the 
right  definition,  which  without  doubt  Eudoxus  or  Euclid  gave, 
in  its  proper  place,  after  the  definition  of  triplicate  ratio 
&c.  in  the  5th  book.  Theon's  definition  is  this ;  a  ratio 
is  said  to  be  compounded  of  ratios  'orctv  ai  tuv  Xoyuv  TtyiXiKor/jne 
i^'  ^ixvtag  7reXA«57Aa(7/«5-.^«5-a!<  7rotai<ri  tivx:  which  Commandine 
thus  translates  :  "  quando  rationvnii  quantitates  inter  sc  multi- 
"  plicate  aliquam  efficiunt  rationem  ;"  that  is,  when  the 
quantities  of  the  ratios  being  multiplied  by  one  another  make  a 
certain  ratio.  Dr.  Wallis  translates  the  word  ^tux^xotjjt:?  "  ra- 
"  tionem  exponentes,"  the  exponents  of  the  ratios :  and  Dr. 
Gregory  renders  the  last  words  of  the  definition  by  "  illius  fa- 
"  cit  quantitatem,"  makes  the  quantity  of  that  ratio :  but  in 
v/hatever  sense  the  "  quantities,"  or  "  exponents  of  the  ra- 
"  tios,"  and  their  multiplication"  be  taken,  the  definition 
win  be  ungeom.etrical  and  useless  :  for  there  can  be  no  multi- 
plication but  by  a  number.  Now  the  quantity  or  exponent  of 
V.  ratio  (according  as  Eutocius  in  his  Comment,  on  prop.  4. 
book  2,  of  Arch,  de  Sph.  ct  Cyl.  and  the  moderns  explain  that 
term)  is  the  number  which  multiplied  into  the  conseqvient  term 
of  a  ratio  produces  the  antecedent,  or,  which  is  the  same  thing, 
the  number  which  arises  by  dividing  the  antecedent  by  the  con- 
sequent ;  but  there  are  many  ratios  such,  that  no  number  can 
arise  from  the  division  of  the  antecedent  by  the  consequent : 
ex.  gr.  the  ratio  of  which  the  diameter  of  a  square  has  to  the  side 
of  it ;  and  the  ratio  which  the  circumference  of  a  circle  has 
to  its  diameter,  and  such  like.  Besides,  that  there  is  not  tlie 
least  mention  made  of  this  definition  in  the  writings  of  Eu- 
clid, Archimedes,  ApoUonius,  or  other  ancients,  though  they 
frequently  make  use  of  compound  ratio :  and  in  this  23d  prop, 
of  the  6th  book,  where  compound  ratio  is  first  mentioned,  there 
is  not  one  word  v/hich  can  relate  to  this  definition,  though 
here,  if  in  any  place,  it  was  necessary  to  be  brought  in ;  but  the 
right  definition  is  expressly  cited  in  these  words :  "  But  the 
"  ratio  of  K  to  M  is  compounded  of  the  ratio  of  K  to  L. 


NOTES.  329 

*  aiid  of  the  ratio  of  L  to  M."  This  definition  therefore  of  Book  VI. 
Theon  is  quite  useless  and  absurd  ;  for  that  Taeon  brought  it  n^"v^Vb^ 
into  the  Elements  can  scarce  be  doubted ;  as  it  is  to  be  found 
in  his  Commentary  upon  Ptolomy's  MiyxXm  Syvr«|(5,  page  62, 
where  he  also  gives  a  childish  explication  of  it  as  agreeing 
only  to  such  ratios  as  can  be  expressed  by  numbers  ;  and  from 
this  place  the  definition  and  explication  have  been  exactly  co- 
pied and  prefixed  to  the  definitions  of  the  6th  book,  as  ap- 
pears from  Hervagius'  edition  :  but  Zambertus  and  Comman- 
dine,  in  their  Latin  translations,  subjom  the  same  to  tnese 
definitions.  Neither  Campanus,  nor  as  it  seems,  the  Arubic 
manuscripts,  from  which  he  made  his  translation,  have  this 
definition.  Clavius,  in  his  observations  upon  it,  rightly  judges 
that  the  definition  of  compound  ratio  might  have  been  made 
after  the  same  manner  in  which  the  definitions  of  duplicate 
and  triplicate  ratio  are  given,  viz.  "  That  as  in  several  magni- 
"  tudes  that  are  continual  proportions,  Euclid  named  the 
"  ratio  of  the  first  to  the  third,  the  duplicate  ratio  of  the 
"  first  to  the  second  ;  and  the  ratio  of  the  first  to  the  fourth, 
"  the  triplicate  ratio  of  the  first  to  the  second,  that  is,  the 
"  ratio  compounded  of  two  or  three  intermediate  ratios  that 
"  are  equal  to  one  another,  and  so  on  ;  so,  in  like  manner,  if 
"  there  be  several  magnitudes  of  the  same  kind,  following  one 
"  another,  which  are  not  continual  proportionals,  the  first  is 
"  said  to  have  to  the  last  the  ratio  compounded  of  all  the  in- 

"  termediate  ratios only  for  this  reason,  that  these  inter- 

"  mediate  ratios  are  interposed  betwixt  the  two  extremes,  viz. 
"  the  first  and  last  miagnitudes  ;  even  as,  in  the  1 0th  definition 
"  of  the  5th  book,  the  ratio  of  the  first  to  the  third  was  called 
"  the  duplicate  ratio,  merely  upon  account  of  two  ratios  be- 
"  ing  interposed  betwixt  the  extremes,  that  are  equal  to  one 
"  another :  so  that  there  is  no  difference  betwdxt  this  com- 
"  pounding  of  ratios,  and  the  duplication  or  triplication  of 
"  them  which  are  defined  in  the  5th  book,  but  that  in  the  du- 
"  plication,  triplication,  &c.  of  ratios,  all  the  intei'posed  ratios 
*'  are  equal  to  one  another ;  whereas,  in  the  compounding  of 
"  ratios,  it  is  not  necessary  that  the  intermediate  ratios  should 
"  be  equal  to  one  another."  Also  Mr.  Edmund  Scarburgh, 
in  his  English  translation  of  the  first  six  books,  page  238,  266, 
expressly  affirms,  that  the  5th  definition  of  the  6th  book,  is 
suppositious,  and  that  the  true  definition  of  compound  ratio  is 
contain jd  in  the  1 0th  definition  of  the  5th  book,  viz.  the  defi- 
nition of  duplicate  ratio,  or  to  be  understood  from  it,  to  wit,  in 
the  same  manner  as  Clavius  has  explained  it  in  the  preceding 
citation.     Yet  these,  and  the  rest  of  the  moderns,  do  notwith- 

3  T 


330  NOTES. 

Book  VI.  standing  retain  this  5th  def.  of  the  6th  book,  and  illustrate  an<J 
^-^'■>^'>«-'  explain  it  by  long  commentaries,  when  they  ought  rather  to 
have  taken  it  quite  away  from  the  Elements. 

For,  by  comparing  def.  5,  book  6,  with  prop.  5,  book  8, 
it  will  clearly  appear  that  this  definition  has  been  put  into  the 
Elements  in  place  of  the  right  one  which  has  been  taken  out 
of  them :  because,  in  prop.  5,  book  8,  it  is  demonstrated  that 
the  plane  number  of  which  the  sides  are  C,  D  has  to  the  plane 
number  of  which  the  sides  are  E,  Z,  (see  Hergavius'  or 
Gregory's  edition,)  the  ratio  which  is  compounded  of  the  ra- 
tios of  their  sides ;  that  is,  of  the  ratios  of  C  to  E,  and  D  to 
Z-:  and  by  def.  5.  book  6.  and  the  explication  given  of  it  by 
all  the  commentutoi's,  the  ratio  which  is  compounded  of  the  ra- 
tios of  C  to  E,  and  D  to  Z,  is  the  ratio  of  the  product  made 
by  tiie  multiplication  of  the  antecedents  C,  D  to  the  product 
of  the  consequents  E,  Z,  that  is,  the  ratio  of  the  plane  number 
of  which  the  sides  are  C,  D  to  the  plane  number  of  which 
the  sides  are  E,  Z.  Wherefore  the  proposition  which  is  the  5th 
def.  of  book  6,  is  the  very  same  with  the  5th  prop,  of  book  8, 
and  therefore  it  ought  necessarily  to  be  cancelled  in  one  of  these 
places  ;  because  it  is  absurd  that  the  same  proposition  should 
stand  as  a  definition  in  one  place  of  the  Elements,  and  be  de- 
monstrated in  another  place  of  them.  Now,  there  is  no  doubt 
that  prop.  5,  book  8,  should  have  a  place  in  the  Elements,  as 
the  same  thing  is  denaonstrated  in  it  concerning  plane  num- 
bers, which  is  demonstrated  in  prop.  23d,  book  6,  of  equiangu- 
lar parallelograms  ;  wherefore  def.  5,  book  6,  ought  not  to  be 
in  the  Elements.  And  from  this  it  is  evident  that  this  definition 
is  not  Euclid's,  but  Theon's,  or  some  other  unskilful  geometer's. 
But  nobody,  as  far  as  I  know,  has  hitherto  shown  the  true 
use  of  compound  ratio,  or  for  what  purpose  it  has  been  in- 
troduced into  geometry:  for  every  proposition  in  wnich 
.  compound  ratio  is  made  use  of,  may  v/ithout  it  be  both  enun- 

ciated and  demonstrated.  Now  the  use  of  compound  ratio 
consists  wholly  in  this,  that  by  means  of  it,  circumlocutions 
may  be  avoided,  and  thereby  propositions  may  be  more  brief- 
ly either  enunciated  or  demonstrated,  or  both  may  be  done, 
for  instance  if  this  23d  proposition  of  the  sxith  book  were  to 
be  enunciated,  without  mentioning  compound  ratio,  it  might 
be  done  as  follows.  If  two  parallelograrns  be  equiangular,  and 
if  as  a  side  of  the  first  to  a  side  of  the  second,  so  any  assumed 
straight  line  be  made  to  a  second  straight  line  ;  and  as  the 
other  side  of  the  first  to  the  other  side  of  the  second,  so  the  se- 
cond sti'aight  line  be  made  to  a  third.  The  first  parallelogram 
is  to  the  second,  as  the  first  straight  line  to  the  third.     And  the 


NOTES.  331 

demonstration  would  be  exactly  the  same  as  we  now  have  it.  Baok  Vi. 
But  the  ancient  geometers,  when  they  observed  tnis  enunciu.-  '^~i'^^'^^>^ 
tion  could  be  made  shorter,  by  giving  a  name  to  the  ratio 
which  the  first  straight  line  has  to  the  last,  by  which  name  the 
intermediate  ratios  might  likewise  be  signified,  of  the  first  to 
the  second,  and  of  the  second  to  the  thiid,  and  so  on,  if  there 
Avere  more  of  them,  they  called  this  ratio  of  the  first  to  the 
last,  the  ratio  compovmded  of  the  ratios  of  the  first  to  the  se- 
cond, and  of  the  second  to  the  third  straight  line :  that  is,  in 
the  present  example,  of  the  ratios  which  are  the  same  with 
the  ratios  of  the  sides,  and  by  this  they  expressed  the  proposi- 
tion more  briefly  thus :  if  there  be  two  equiangular  parallel- 
ograms, they  have  to  one  another  the  ratio  which  is  the 
same  with  that  which  is  compounded  of  ratios  that  are  the 
same  with  the  ratios  of  the  sides.  Which  is  shorter  than  the 
preceding  enunciation,  but  has  precisely  the  same  meaning. 
Or  yet  shorter  thus  :  equiangular  parallelograms  have  to  one 
another  the  ratio  which  is  the  same  with  that  which  is  com- 
pounded of  the  ratios  of  their  sides.  And  these  two  enuncia- 
tions, the  first  especially,  agree  to  the  demonstration  which  is 
now  in  the  Greek.  The  proposition  may  be  more  briefly  de- 
monstrated, as  Candulla  does,  thus  :  let  ABCD,  CEJ:  G  be 
two  equiangular  parallelograms,  and  complete  the  parallelo- 
gram CDHG  ;  then,  because  there  are  three  parallelograms 
AC,  CH,  CF,  the  first  AC  (by  the  definilion  of  compound 
ratio)  has  to  the  third  CF,  the  ratio     A  D  H 

which  is  compounded  of  the  ratio  of 
the  first  AC  to  the  second  CH,  and  of 
the  ratio  of  CH,  to  the  third  CF ;  but  B 
the  paralleogram  AC  is  to  the  pa- 
rallelogram CH,  as  the  straight  line 
BC  to  CG ;  and  the  parallelogram 

CH  is  to  CF,   as  the  straight  line  E    .  F 

CD  is  to  CE ;  therefore  the  parallelogram  AC  has  to  CF  the 
ratio  which  is  compounded  of  ratios  that  are  the  same  v.ith  the 
ratios  of  the  sides.  And  to  this  demonstration  agrees  the  enun- 
ciation which  is  at  present  in  the  text,  viz.  Equiangular  parallel- 
ograms have  to  one  another  the  ratio  which  is  compovmded  of 
the  ratios  of  the  sides  :  for  the  vulgar  reading,  "  wiiich  is  com- 
"  pounded  of  their  sides,"  is  absurd.  But,  in  this  edition,  vie 
have  kept  the  demonstration  which  is  in  the  Greek  text,  though 
not  so  short  as  Candalla's  ;  because  the  way  of  finding  the  ratio 
which  is  compounded  of  the  ratios  of  the  sides,  that  is,  of  find- 
ing the  ratio  of  the  parallelograms,  is  shown  in  that,  but  not  in 
Candalla's  demonstration  ;  whereby  beginners  may  learn,  in  likg 


1 

c 

332  NOTES. 

Book  VI.  cases,  how  to  find  the  ratio  which  is  compounded  of  two  or  more 

v^'"'<'">*^  given  ratios. 

From  what  has  been  said,  it  may  be  observed,  that  in  any 
magnitudes  whatever  of  the  same  kind  A,  B,  C,  D,  Sec  the 
ratio  compounded  of  the  ratios  of  the  first  to  the  second,  of 
the  second  to  the  third,  and  so  on  to  the  last,  is  only  a  name 
or  expression  by  which  the  ratio  which  the  first  A  has  to  the 
last  D  is  signified,  and  by  wliich  at  the  same  time  the  ratios  of 
all  the  magnitudes  A  to  B,  B  to  C,  C  to  D  from  the  first  to 
the  last,  to  one  another,  whether  they  be  the  same,  or  be  not 
the  same,  are  indicated ;  as  in  magnitudes  which  are  continual 
proportionals  A,  B,  C,  D,  &.c.  the  duplicate  ratio  of  the  first 
to  the  second  is  only  a  name,  or  expression  by  which  the  ratio 
of  the  first  A  to  the  third  C  is  signified,  and  by  which,  at  the 
same  time,  is  shown  that  there  are  two  ratios  of  the  magni- 
tudes, from  the  first  to  the  last,  viz.  of  the  first  A  to  the  se- 
cond B,  and  of  the  second  B  to  the  third  or  last  C,  which  are 
the  same  with  one  another ;  and  the  triplicate  ratio  of  the 
first  to  the  second  is  a  nam^  or  expression  by  which  the  ratio 
of  the  first  A  to  the  fourth  D  is  signified,  and  by  which,  at  the 
same  time  is  snown  that  there  are  three  ratios  of  the  magni- 
tudes fiom  the  first  to  the  last,  viz.  of  the  first  A  to  the  se- 
cond B,  and  of  B  to  the  third  C,  and  of  C  to  the  fourth  or 
last  D,  which  are  all  the  same  with  one  another ;  and  so  in 
the  case  of  any  other  multipiicate  ratios.  And  that  this  is 
the  right  explication  of  the  meaning  of  these  ratios  is  plain 
from  tne  definitions  of  duplicate  and  triplicate  ratio  in  which 
Euclid  makes  use  of  the  Avord  .-viici,  is  said  to  be,  or  is  called; 
which  V/ord,  he,  no  doubt,  made  use  of  also  in  the  definition 
of  compound  ratio,  which  Theon,  or  some  other,  has  expung- 
ed from  the  Elements  ;  for  the  very  same  word  is  still  retained 
in  the  wrong  definition  of  compound  ratio,  which  is  now  the 
5th  of  the  6th  book  :  but  in  the  citation  of  these  definitions  it 
is  some  times  retained,  as  in  the  demonstration  of  prop.  19, 
book  6,  "  the  first  is  sajd  to  have,  '^z-hv  asy6t«(,  to  the  third  the 
"  duplicate  ratio,"  &c,  wi:ich  is  wrong  translated  by  Comman- 
dine  and  others,  "  has"  instead  of  "  is  said  to  have  :"  and 
sometemes  it  is  left  out,  as  in  the  demonstration  of  prop.  33. 
of  the  1 1th  book,  in  which  Ave  find  "  the  first  has,  '3;%j«,  to  the 
"  third  the  triplicate  ratio  ;'*  but  without  doubt  's^  <,  "  has," 
in  this  place  signifies  the  same  as  's;v^^v  Xiyirai^  is  said  to  have: 
so  likewise  in  prop.  23,  B.  6,  we  find  this  citation,  "  but  the 
"  ratio  of  K  to  M  is  compounded,  c-vyxitrxi,  of  the  ratio  of 
"  K  to  L,  and  the  ratio  of  L  to  M,"  which  is  a  shorter  way  of 
expressing  the  sa.nie  thing,  which,  according  to  the  d-efinition. 


NOTES.  333 

ought  to  have  been  expressed  by  o-!;v>6««-.^«<  AsysTai,  is  said  so  be  Book  VI. 
compounded.  Si^'N'^^/ 

From  these  remarks,  together  with  the  propositions  subjoined 
to  tne  5th  book,  aii  tnat  is  found  concerning  compound  ratio, 
either  in  the  ancient  or  modern  geometers,  may  be  understood 
and  explained. 


PROP.  XXIV.     B.  VI. 

It  seems  that  some  unskilful  editor  has  made  up  this  demon- 
stration as  we  now  have  it,  out  of  two  others ;  one  of  whici-  may 
be  made  from  the  2d  prop,  and  the  otner  from  the  4th  or  tuis 
book  :  for  after  he  has,  from  the  2d  of  this  book,  and  compo- 
sition and  permutation,  demonstrated  that  the  sides  about  the 
angle  common  to  the  two  parallelograms  are  propoi'tionals,  he 
might  nave  immeciia.tely  concluded  that  the  sides  about  the  other 
equal  angles  were  proportionals,  viz.  from  prop  34,  B.  1,  and 
prop.  7,  book  5.  This  he  does  not,  but  proceeds  to  show  that 
the  triangles  and  parallelograms  are  equiangular  ;  and  in  a  te- 
dious way,  by  help  of  prop.  4,  of  this  book,  and  the  22d  of 
book  5,  deduces  the  same  conclusion :  from  which  it  is  plain 
that  this  ill  composed  demonstration  is  not  Euclid's  :  these  su- 
perfluous things  are  now  left  out,  and  a  more  simple  demonstra- 
tion is  given  from  the  4th  prop,  of  this  book,  the  same  which 
is  in  tne  translation  from  the  Arabic,  by  the  help  of  the  2d  prop, 
and  composition  ;  but  in  this  the  author  neglects  permutation, 
and  does  not  show  the  parallelograms  to  be  equiangular,  as  is 
proper  to  do  for  the  sake  of  beginners. 


PROP.  XXV.     B.  VI. 

It  is  very  evident  that  the  demon sti^ation  which  Euclid  had 
given  of  this  proposition  has  been  vitiated  by  some  unskilful 
hand :  for,  after  this  editor  had  demonstrated  that  "  as  the 
"  rectilineal  figure  ABC  is  to  the  rectilineal  KGH,  so  is  the 
"  parallelogram  BE  to  the  parallelogram  EF  ;"  nothing  more 
should  have  been"  added  but  this,  "  and  the  rectilineal  figure 
"  ABC  is  equal  to  the  parallelogram  BE:  therefore" the  recti- 
"  lineal  KGH  is  equal  to  the  parallelogram  EF,"  viz.  from 
prop.  14,  book  5.  But  betwixt  these  two  sentences  he  has  in- 
serted this  ;  ''  wherefore  by  permutation,  as  the  rectilineal  fi- 
"  gure  ABC  to  the  parallelogram  BE,  so  is  the  rectilineal  KGH 


354  NOTES. 

^Book  VI.  "  to  the  parallelogram  EF  ;"  by  which,  it  is  plain,  he  thought 
^i^'V''^^  it  was  not  so  evident  to  conclude  tliat  the  second  of  four  pro- 
portionals is  equal  to  the  fourth  from  the  equality  of  the  first 
and  third,  which  is  a  thing  demonstrated  in  the  l4th  prop,  of 
B.  5,  as  to  conclude  that  the  third  is  equal  to  the  fourth,  from 
the  equality  of  the  first  and  stcond,  wiiich  is  no  where  demon- 
strated in  the  Elements  as  Ave  now  have  them  :  but  though  this 
proposition,  viz.  the  third  of  four  proportionals  is  equal  to  the 
fourth,  if  the  third  be  equal  to  the  second,  had  been  given  in 
the  Elements  by  Euclid,  as  very  probably  it  was,  yet  he  would 
not  have  made  use  of  it  in  this  place  ;  because,  as  was  said  the 
conclusion  could  have  been  immediately  deduced  without  this 
superfluous  step  by  permutation  :  this  v/e  have  shown  at  the 
greater  length  both  because  it  affords  a  certain  proof  of  the 
Adtiation  of  the  text  of  Euclid  ;  for  the  very  same  blunder  is 
found  twice  in  the  Greek  text  of  prop.  23,  book  1 1,  and  twice 
in  prop.  2,  B.  12,  and  in  the  5,  11,  12,  and  18th  of  that  book ; 
in  which  places  of  book  12,  except  the  last  of  them,  it  is  rightly 
left  out  in  the  Oxford  edition  of  Commandine's  translation  ; 
and  also  that  geometers  may  beware  of  making  use  of  permu- 
tation in  the  like  cases  ;  for  the  moderns  not  unfrequently  com- 
mit this  mistake,  and  among  others  Commandine  himself  in  his 
commentary  on  prop.  5,  book  3,  p.  6,  b.  of  Pappus  Alexandri- 
nus,  and  in  other  places  :  the  vulgar  notion  of  proportionals 
has,  it  seems,  pre-occupied  many  so  much,  that  they  do  not  suf- 
ficiently understand  the  true  nature  of  them. 

Besides,  though  the  rectilineal  figure  ABC,  to  which  another 
is  to  be  made  similar,  may  be  of  any  kind  whatever  ;  yet  in  the 
demonstration  the  Greek  text  has  "  triangle"  instead  of  "  recti- 
"  lineal  figure,"  which  error  is  corrected  in  the  above  named 
Oxford  edition. 

PROP.  XXVII.     B.  VI. 

The  second  case  of  this  has  '«aa»?,  otherwise,  prefixed  to 
it,  as  if  it  was  a  different  demonstration,  which  probably  has 
been  done  by  some  unskilful  librarian.  Dr.  Gregory  has  rightly 
left  it  out :  the  scheme  of  this  second  case  ougiit  to  be  marked 
with  the  same  letters  of  the  alphabet  which  are  in  the  scheme 
of  the  first,  as  is  now  done. 


PROP.  XXVIII  and  XXIX.     B.  VI. 

These  two  problems,  to  the  first  of  which  the  27th  prop,  iiv 
iiecessai-y,  are  the  most  general  and  useful  in  all  the  Elements, 


NOTES.  3*35 

and  are  most  frequently  made  use  of  by  the  ancient  gometers  Book  VI. 
in  the  solution  of  other  problems  ;  and  therefore  are  very  igno-  *<^^"v-Xi,^ 
rantly  left  out  by  Tacquet  and  Dechales  in  their  editions  of  the 
Elements,  who  pretend  that  they  are  scarce  of  any  use.  The 
cases  of  these  pi'oblems,  wherein  it  is  required  to  apply  a  rect- 
angle which  shall  be  equal  to  a  given  square,  to  a  given  straight 
line,  either  deficient  or  exceeding  by  a  square  ;  as  also  to  apply 
a  rectangle  which  shall  be  equal  to  another  given,  to  a  given 
straight  line,  deficient  or  exceeding  by  a  square,  are  very  often 
made  use  of  by  geometers.  And,  on  this  account,  it  is  thought 
proper,  for  the  sake  of  beginners,  to  give  their  constructions  as 
follows : 

1 .  To  apply  a  rectangle  which  shall  be  equal  to  a  given  square, 
to  a  given  straight  line,  deficient  by  a  square  :  but  the  given 
square  must  not  be  greater  than  that  upon  the  half  of  the  given 
line. 


I 

n^ 

D       /G 

/ 

c 

K 


B 


Let  AB  be  the  given  straight  line,  and  let  the  square  upon 
the  given  straight  line  C  be  that  to  which  the  rectangle  to  be  ap- 
plied must  be  equal,  and  this  square  by  the  determination,  is  not 
greater  than  that  upon  half  of  the  straight  line  AB. 

Bisect  AB  in  D,  and  if  the  square  upon  AD  be  equal  to 
the  square  upon  C,  the  thing  required  is  done  :  but  if  it  be  not 
equal  to  it,  AD  must  be 

greater  than  C,  according     L  H 

to  the  determination :  draw 
DE  at  right  angles  to  AB 
and  make  it  equal  to  C :  A 
produce  ED  to  F,  so  that 
EF  be  equal  to  AD  or  DB 
and  from  the  centre  E,  at 
the  distance  EF,  describe  a 
circle  meeting  AB  in  G,  E 

and  upon  GB  describe  the  square  GBKH,  and  complete  the 
rectangle  AGHE ;  also  join  EG.  And  because  AB  is  bisected 
in  D,  the  rectangle  AG  GB  together  with  the  square  of  DG 
is  equal  a  to  (the  square  of  DB,  that  is,  of  EF  or  EG,  that  is,  a  5.  2. 
to)  the  squares  of  ED,  DG  :  take  away  the  square  of  DG 
from  each  of  these  equals  ;  therefore  the  remaining  rectangle 
AG,  GB,  is  equal  to  the  square  of  ED,  that  is,  of  C  :  but  the 
rectangle  AG,  GB  is  the  rectangle  AH,  because  GH  is  equal 
to  GB ;  therefore  the  rectangle  AH  is  equal  to  the  given  square 
upon  the  straight  line  C.  Wherefore  the  rectangle  AH,  equal 
to  the  given  square  upon  C,  has  been  applied  to  the  given 


336  NOTES. 

Book  VI.  straight  line  AB,  deficient  by  the  squai^e  GK.      Which  was  to 
^^-^"^^"^^fm^  be  done. 

2.  To  apply  a  rectangle  Avhich  shall  be  equal  to  a  given 
squtU'e,  to  a  given  straight  line,  exceeding  by  a  square. 

Let  AB  be  the  given  straight  line,  and  let  the  square  upon 
the  given  straight  line  C  be  that  to  M^hich  the  rectangle  to  be 
applied  must  be  equal. 

Bisect  AB  in  D,  and  dravi^  BE  at  right  angles  to  it,  so  that 
BE  be  equal  to  C  ;  and  having  joined  DE,  from  the  centre  D 
at  the  distance  DE  describe  a  circle  meeting  AB  pi'oduced  in 
G  ;  upon  BG  describe  the  square 
BGHK,  and  complete  the  rc;ct- 
angle  AGHL.   And  because  AB 
is  bisected  in  D,  and  produced 
to   G,    the   rectangle    AG,  GB 
together  with  the  square  of  DB 
a  6.  2.     is  equal  a  to  (the  square  of  DG 

or  DE,  that  is,  to)  the  squares  F 

of  EB,  BD.  From  each  of  these 

equals  take  the  square  of  DB  ;  C 

therefore  the  remaining  rectangle  AG,  GB  is  equal  to  the 
square  of  BE,  that  is,  to  the  square  upon  C.  But  the  rectan- 
gle AG,  GB  is  the  rectangle  AH,  because  GH  is  equal  to  GB : 
therefore  the  rectangle  AH  is  equal  to  the  square  upon  C. 
Wherefore  the  rectangle  AH,  equal  to  the  given  square  upon 
C,  has  been  applied  to  the  given  straight  line  AB,  exceeding 
by  the  square  GK.     Which  was  to  be  done. 

3.  To  apply  a  rectangle  to  a  given  straight  line  which  shall 
be  equal  to  a  given  rectangle,  and  be  deficient  by  a  square.  But 
the  given  rectangle  must  not  be  greater  than  the  square  upon 
the  half  of  the  given  straight  line. 

Let  AB  be  the  given  straight  line,  and  let  the  given  rectan- 
gle be  that  which  is  contained  by  the  straight  lines  C,  D  Av.iich 
is  not  greater  than  the  square  upon  tlie  half  of  AB  ;  it  is  re- 
quired to  apply  to  AB  a  rectangle  equal  to  the  rectangle  C,  D, 
deficient  by  a  square. 

Draw  AE,  BF  at  right  angles  to  AB,  upon  the  same  side  of  it, 
and  make  AE  equal  to  C,  and  BF  to  D  :  join  EF,  and  bisect  it 
in  G  ;  and  from  the  centre  G,  at  tiie  cUatance  GE,  describe  a 
circle  mes^ting  AE  again  in  H  ;  join  HF,  and  draw  GK  parallel 
to  it,  and  GL  parallel  to  AE,  meeting  AB  in  L. 


NOTES. 


337 


Because  the  angle  EHF  in  a  semicircle  is  equal  to  the  right  Book  VI. 
angle  EAB,  AB  and  HF  are  parallels,  and  AH  and  BF  are  ^-^'■^'"^i^ 
parallels  ;  wherefore  AH  is  equal  to  BF,  and  the  rectangle 
EA,   AH   equal   to   the   rectangle   EA,   BF,  that  is,  to  the 
rectangle  C,  D  :   and  because  EG,  GF  are  equal  to  one  another, 
and  AE,  LG,  BF  parallels  :  therefore  AL  and  LB  are  equal ; 
also  EK  is  equal  to  KH  a,  and  the  rectangle  C,  D  from  the  a  3.  3. 
determination,  is  not  greater  than  the  square  of  AL  the  half 
of  AB  ;  wherefore  the  rectangle  EA,  AH  is  not  gi'eater  than 
the   square  of  AL,  that  is  of    KG  :  add  to  each  the  square 
of  KE  ;  therefore  the  square  b  of  AK  is  not  greater  than  the  b  6.  2. 
squares  of  EK,  KG,  that  is, 

^  C 


than  the  squre  of  EG  ;  and 
consequently  the  straight  line 
AK  or  GL  is  not  greater 
than  GE.  Now,  if  GE  be 
equal  to  GL,  the  circle  EHF 
touches  AB  in  L,  and  there. 
fore  the  square  of  AL  is  c 
equal  to  tiie  rectangle  EA, 
AH,  that  is,  to  the  given  rect- 
angle C,  D  ;  and  that  wiiich 
was  required  is  done  :  but  if 
EG,  GL  be  unequal,  EG 
must  be    the    greater :    and 


c  36. 


therefore  the  circle  EHF  cuts  the  straight  line  AB  ;  let  it  cut  it 

in  the  points  M,  N,  and  upon  NB  describe  the  square  NBOP, 

and  complete  the  rectangle  ANPQ  :  because  ML  is  equal  to  dj.  3.  3. 

LN,  and  it  has  been  proved  that  AL  is  equal  to  LB  ;  therefore 

AM  is  equal  to  NB,  and  the  rectangle  AN,  NB  equal  to  the 

rectangle  N  A,  AM,  that  is,  to  the  rectangle  e  E  A,  AH,  or  the  g  q^^  gg. 

rectangle  C,  D  :  but  the  rectangle  AN,  NB  is  the  rectangle  3. 

AP,  because  PN  is  equal  to  NB  :  therefore  the  rectangle  AP 

is  equal  to  the  rectangle  C,  D  ;  and  the  rectangle  AP  equal  to 

the  given  rectangle  C,  D  has  been  applied  to  the  given  straight 

line  AB,  deficient  by  the  square  BP.     Wiiich  Vv^as  to  be  done. 

4.  To  apply  a  rectangle  to  a  given  straight  line  that  shall  be 
equal  to  a  given  rectangle,  exceeding  by  a  square 

Let  AB  be  the  given  straight  line,  and  the  rectangle  C,  D  the 
given  rectangle,  it  is  required  to  apply  a  rectangle  to  AB  equal 
to  CD,  exceeding  by  a  square. 

Draw  AE,  BF  at  right  angles  to  AB,  on  the  contrary  sides 
of  it,  and  make  AE  equal  to  C,  and  BF  equal  to  D  :  join 
EF,  and  bisect  it  in  G  ;  and  from  the  centre  G,  at  the  distance 

^  U 


338 


NOTES. 


Book  VI  GE,  describe  a  circle  meeting  AE  again  in  H  ;  join  HF,  and 
v.^^>^'^^  draw  GL  parallel  to  AE  ; 
let  the  circle  meet  AB  pro- 
duced in  M,  N,  and  upon 
BN  describe  the  square 
NBOP,  and  complete  the 
rectangle  ANFQ  :  because 
the  angle  EHl-  in  a  semi- 
circk  is  equal  to  the  right 
angle  EAB,  AB  cUid  HF 
are  parallels,  and  therefore 
AH  and  BF  are  eqUcil,  and 
the  rectangle  E  A,  AH  equal 
to  the  rectangle  E  A,BF,that 
is,  to  the  rectangle  C,  D  :  and 

because  ML  is  equal  to  LN,  and  AL  to  LB,  therefoi'e  MA  is 
equal  to  BN,  and  the  rectangle  AN,  NB  to  MA,  AN,  that  is, 
a  35.  3.  a  to  the  rectangle  EA,  AH,  or  the  rectangle  C,  D  :  therefore 
the  rectangle  AN,  NB,  that  is,  AP,  is  equal  to  the  rectangle 
C,  D  ;  and  to  the  given  straight  line  AB  the  rectangle  AP  has 
been  applied  equal  to  the  given  rectangle  C,  D,  exceeding  by 
the  square  BP.     Which  was  to  be  done. 

Willebrordus  Snellius  was  the  first,  as  far  as  I  know,  who 
gave  these  constructions  of  the  3d  and  4th  problems  in  his  Ap- 
pollonius  Batavus  :  and  afterwards  the  learned  Dr.  Halley  gave 
them  in  the  Scholium  of  the  18th  prop,  of  the  8th  book  of  Ap- 
polonius's  conies  restored  by  him. 

The  3d  problem  is  otherwise  enunciated  thus :  To  cut  a 
given  straight  line  AB  in  the  point  N,  so  as  to  make  the  rect- 
angle AN,  NB  equal  to  a  given  space  :  oi",  which  is  the 
same  thing,  having  given  AB  the  sum  of  the  sides  of  a  rect- 
angle, and  the  magnitude  of  it  being  likewise  given,  to  find  its 
sides. 

And  the  fourth  problem  is  the  same  with  this.  To  find  a  point 
N  in  the  given  straight  line  AB  produced,  so  as  to  make  the 
rectangle  AN,  NB  equal  to  a  given  space  :  or,  wiiich  is  the 
same  thing,  having  given  AB  the  difterence  of  the  sides  of  a 
rectangle,  and  the  magnitude  of  it,  to  find  the  sides. 


NOTES. 


PROP.  XXXI.     B.  VI. 


In  the  demonstration  of  this,  the  inversion  of  proportionals  is 
twice  neglected,'  and  is  now  added,  that  the  conclusion  m^y  be 
legitimately  made  by  help  of  the  24th  prop,  of  B.  5.  as  Clavius 
had  done. 


PROP.  XXXII.     B.  VI. 

The  enunciation  of  the  preceding  26th  prop,  is  not  general 
enough  ;  because  not  only  two  simii^ir  paralleiogranis  tiuit  have 
an  angle  common  to  both,  are  about  the  same  cliumeter  ;  but 
likewise  two  similar  parallelograms  that  have  vertically  opposite 
angles,  have  their  diameters  in  the  same  straight  line :  but  ti.ere 
seems  to  have  been  another,  and  that  a  direct  demonstration  of 
these  cases,  to  which  this  32d  proposition  was  needful :  and  the 
32d  may  be  otherwise  and  something  more  briefly  demonstrated 
as  follows : 


PROP.  XXXII.    B.  VI. 


If  two  triangles  which  have  two  sides  of  the  one,  &c. 
Let  Gx\F,  HFC  be  two  triangles  which  have  two  sides  AG, 
GF,  proportional  to  the  two  sides  fH,  HC,  viz.  AG  to  GF,  as 


FH  to  HC  ;  and  let  AG  be  paral- 
lel to  FH,  and  GF  to  HC  ;  AF 
and  FC  are  in  a  straight  line. 

Draw  CK  parallel  a  to  FH,  and 
let  it  meet  GF  produced  in  K ; 
because  AG,  KC  are  each  of  them 
parallel  to  FH,  they  are  parallel  b 
to  one  another,  and  therefore  the 
alternate  angles  AGF,  FKC  are 


G 


F 


D 


;H 


a  31.1. 


b  30.  1. 


B 


K 


14.  1. 


equal :  and  AG  is  to  GF,  as  (FH  to  HC,  that  is  c)  CK  to  KF  ;  c  34. 1 
wherefore  the  triangles  AGF,  CKF  are  equiangular  d,  and  the  d  6.  6. 
angle  AFG  equal  to  the  angle  CFK  :  but  GFK  is  a  straight  line, 
therefore  AF  and  FC  are  in  a  straight  linee. 

The  26th  prop,  is  demonstrated  from  the  32d,  as  follows  : 
If  two  similar  and  similarly  placed  parallelograms  have  an 
angle  common  to  both,  or  vertically  opposite  angles  ;  their  di- 
ameters are  in  the  same  straight  line. 

First,  Let  the  parallelogram-j  ABCD,  AEFG  have  the  angle 
BAD  common  to  both,  and  be  similar,  and  similarly  placed  ; 
ABCD,  AEFG  are  about  the  same  diameter. 


340 


NOTES. 


a  Cor. 
5, 


Book  VI.       Produce  EF,  GF,  to  H,  K,  and  join  FA,  FG  :   then  be- 

y-^^^^^"*-^  caust   tiie  parallelograms   ABCD,    AE,FG   are   similar,   DA 

is  to  AB,  as  GA  to  AE  :  where-       AG  D 

19.  fore  the  remainder  DG  is  a  to  the 

remainder  EB,  as  G  A  to  AE :  but 

DG  is  equal  to  FH,  EB  to  HC,       E  [ ^i^ j  h 

and  AE  to  GF  :  therefore  as  FH 
to  HC,  so  is  AG  to  GF  ;  and 
FH,  HC  are  parallel  to  AG,  GF  ; 
and  the  triangles  AGF,  FHC  are 
joined  at  one  angle,  in  the  point 
6.      F  ;  wherefore  Ai' ,  FC  are  in  the  same  straight  line  b. 

Next,  Let  the  parallelograms  KFHC,GFEA,  which  are  simi- 
lar and  similarly  placed,  have  their  angles  KFH,  GFE  vertically 
opposite ;  their  diameters  AF,  FC  are  in  the  same  straight  line. 
Because  AG,  GF  are  parallel  to  FH,  HC  ;  and  that  AG 
is  to  GF,  as  FH  to  HC:  therefore  AF,  FC  ai'e  in  the  same 
straight  line  b. 


A 

G 

E 

\ 

r 

\ 

B 


K 


PROP.  XXXHI.     B.  VL 

The  words  "  because  they  are  at  the  centre,"  are  left  out,  as 
the  addition  of  some  unskilful  hand. 

In  the  Greek,  as  also  in  the  Latin  translation,  the  words 
«  irv)(,h  "  any  whatever,"  are  left  out  in  the  demonstration  of 
both  parts  of  the  proposition,  and  are  now  added  as  quite  neces- 
sary ;  and  in  the  demonstration  of  the  second  part,  where  the 
triangle  BGC  is  proved  to  be  equal  to  CGK,  the  illative  par- 
ticle ci£»  in  the  Greek  text  ought  to  be  omitted. 

The  second  part  of  tl;e  proposition  is  an  addition  of  Thecn's, 
as  he  tells  us  in  his  commentary  on  Ptolemy's  Miyd?,-,;  I^wTa^ic^ 
p.  50. 


PROP.  B.C.  D.     B.VL 

These  three  propositions  are  added,  because  they  are  fre- 
quently made  use  ol  by  geometers. 


NOTES. 


DEF.  IX.  and  XL     B.  XI. 


THE  similitude  of  plane  figures  is  defined  from  the  equality 
of  their  angles,  and  tiie  proportionality  of  tiie  siaes  about  the 
equal  angles ;  for  from  the  proportionality  of  the  sidv^s  oniy,  or 
ouiy  from  tne  equality  of  the  angles,  the  similitude  of  the  figures 
does  not  follow,  except  in  the  case  wnen  the  figures  are  trian- 
gles :  the  similar  position  of  the  sides  whicli  contain  the  figures, 
to  one  another,  depenuing  partly  upon  eacr.  of  these :  and,  for 
the  same  reason,  those  are  similar  soiid  figures  which  have  all 
their  solid  angles  equal,  each  to  each,  and  are  contained  by  the 
same  number  of  similar  plane  figures :  for  there  are  some  solid 
figures  contained  by  similar  plane  figures,  of  the  same  number, 
and  even  of  the  same  magnitude,  that  are  neither  similar  nor 
equal,  as  shall  be  demonstrated  after  the  notes  on  the  10th  defi- 
nition :  upon  this  account  it  was  necessary  to  amend  the  defini- 
tion of  similar  solid  figures,  and  to  place  the  definition  of  a  solid 
angle  before  it:  and  from  this  and  the  10th  definition,  it  is  suf- 
ficiently plain  how  much  the  Elements  have  been  spoiled  by 
unskilful  editors. 


DEF.  X.     B.  XI. 


Since  the  meaning  of  the  word  "  equal"  is  known  and 
established  before  it  comes  to  be  used  in  this  definition  ; 
therefore  the  proposition  which  is  the  10th  definition  of  this 
book,  is  a  theorem,  the  truth  or  falsehood  of  which  ought  to 
be  demonstrated,  not  assumed  ;  so  that  Theon,  or  some 
other  editor,  has  ignorantly  turned  a  theorem  which  ought 
to  be  demonstrated  into  this  10th  definition:  that  figures  are 
similar,  ought  to  be  proved  from  the  definition  of  similar 
figures;  that  they  are  equal  ought  to  be  demonstrated  from 
the  axiom,  "  Magnitudes  that  wholly  coincide,  are  equal 
"to  one  another;"  or  from  prop.  A.  of  book  5,  or  the  9th 
prop,  or  the  14th  of  the  same  book,  from  one  of  which  the 
equality  of  all  kind  of  figures  must  ultimately  be  deduced. 
In  the  preceding  books,  Euclid  has  given  no  definition  of 
equal  figures,  and  it  is  certain  he  did  not  give  this:  for  v/hat  is 


342 


NOTES. 


Book  XI.  called   the    1st  def.  of  the   5d  book,  is  really  a  theorem  in 

v^''^'^^*^  Wiiich  these  circles  are  said  to  be  equal,  that  have  the  straight 
lines  from  their  centres  to  the  circumferences  equal,  which  is 
plain,  from  the  defiration  of  a  circle;  and  thertfoie  has  by- 
some  editor  been  improperly  placed  among  the  dennidoris.  The 
equality  of  figures  ought  not  to  be  defined,  but  demonstrated : 
therefore,  though  it  were  true,  that  solid  figures  contained  by 
the  same  nunaber  of  similar  and  equal  plane  figures  are  equal 
to  one  another,  yet  he  would  justly  deserve  to  be  blamed  wi^,o 
would  make  a  definition  of  this  proposition  w!,ich  ought  to  be 
demonstrated.  But  if  this  pi'oposidon  be  not  true,  must  it  not 
be  confessed,  that  geometers  have,  for  these  thirteen  hundred 
years,  been  mistaken  in  tiiis  elementary  matter  ?  And  tl  is  should 
teach  us  modesty,  and  to  acknowledge  how  little,  through  the 
weakness  of  our  minds,  we  are  able  to  prevent  mistakes  even  m 
the  principles  of  sciences  which  are  justly  reckoned  amongst 
the  most  certain ;  for  that  the  proposition  is  not  universally  true, 
can  be  shewn  by  many  examples;  the  following  is  sufficient. 
Let  there  be  any   plane  rectilineal  figure,  as  the  triangle 

■A  12.  11.  ABC,  and  from  a  point  D  witiiin  it  draw  a  the  straight  Ijne 
DE  at  right  angles  to  the  plane  ABC ;  in  DE  take  DE,  DP 
equal  to  one  another,  upon  the  opposite  sides  of  the  plane, 
and  let  G  be  any  point  in  EF  ;  join  DA,  DB,  DC ;  EA 
EB,  EC ;  FA,  FB,  FC ;  G A,  GB,  GC :  because  the  straight 
line  EDF  is  at  right  angles  to  the  plane  ABC,  it  makes  iight 
angles  Avith  DA,  DB,  DC  which  it  meets  in  that  plane  ;  and 
in  the  triangles  EDB,  FDB,  ED  and  DB  are  equal  to  FD  and 
DB,  each  to  each,   and  they  contain  right  angles  ;  therefore 

b  4.  1.  ■  the  base  EB  is  equal  b 
to  the  base  FB;  in  the 
same  manner  EA  is  e- 
qual  to  FA,  and  EC  to 
FC :  and  in  the  triangles 
EBA,  FBA,  EB,  BA 
ai'c  equal  to  FB,  BA, 
and  the  base  EA  is  e- 
qual  to  the  base  FA ; 
wherefore      the      angle 

c  8.  1.       EBA   is   equalc    to  the  ^ 

anele  FBA,  and  the  tri-      „   . '^--"^      I   ■  . "~         C 

angle    EBA    equal  b   to 
the  triangle   FBA,    and 
the  other  angles  equal  to 
r  4.  6.   the  other  angles ;  there-  p 

d<  I. def.  fore  these   triangles  are 
^    ^-     similar  d;  in  the  same  manner  the  triangle  EBC  is  similar  to 


NOTES.  343 

the  triangle  FBC,  and  the  triangle  EAC  to  FAC  ;  therefore  Book  XI. 
there  are  two  solid  figures,  each  of  which  is  contained  by  six  v.^'"v->»/ 
triangles,  one  of  them  by  three  tri^^ngles,  the  common  vertex 
of  which  is  the  pomt  G,  and  their  bases  the  straight  lines  AB, 
BC,  CA,  and  by  three  other  triangles  the  common  vertex 
of  which  is  the  point  E,  and  their  bases  the  same  lines  AB, 
BC,  C  A ;  the  otner  solid  is  contained  by  the  same  three  tri- 
angles the  common  vertex  of  which  is  G,  and  their  bases  AB, 
BC,  CA  ;  and  by  three  other  triangles  of  which  the  common 
vertex  is  the  point  F,  and  their  bases  the  same  straight  lines 
AB,  BC,  CA:  now  the  three  triangles  GAB,  GBC,  GCA 
are  common  to  both  solids,  and  the  three  othei-s  EAB,  EBC, 
EC  A,  of  the  first  solid  have  been  shown  equal  and  similar  to  the 
three  others  FAB,  FBC,  FCA  of  the  other  solid,  each  to  each  ; 
therefore  these  two  solids  are  contained  by  the  same  number  of 
equal  and  similar  planes  :  but  that  they  are  not  equal  is  mani- 
fest, because  the  first  of  them  is  contained  in  the  other  :  there- 
fore it  is  not  univers  Jly  true  tiiat  solids  are  equal  which  are 
contained  by  the  same  number  of  equal  and  similar  planes. 

Cor.  From  this  it  appears  that  two  unequal  solid  angles  may 
be  contained  by  the  same  number  of  equal  plane  angles. 

For  the  solid  angle  at  B,  which  is  contained  by  the  four  plane 
angles  EBA,  EBC,  GBA,  GBC  is  not  equal  to  the  solid  angle 
at  the  same  point  B  which  is  contained  by  the  four  plane  angles 
FBA,  FBC,  GBA,  GBC  ;  for  this  last  contains  the  other  :  and 
each  of  them  is  contained  by  four  plane  angles,  which  are  equal 
to  one  another,  each  to  each,  or  are  the  self  same  ;  as  has  been 
proved  :  and  indeed  there  may  be  innunierable  solid  angles  all 
unequal  to  one  another,  wiiich  are  each  of  them  contained  by 
plane  angles  that  are  equal  to  one  another,  each  to  each  :  it  is 
likewise  manifest  that  the  before-mentioned  solids  are  not  simi- 
lar, since  their  solid  angles  are  not  all  equal. 

And  that  there  may  be  innumerable  solid  angles  all  unequal 
to  one  another,  which  are  each  of  them  contained  by  the  same 
plane  angles  disposed  in  the  same  order,  will  be  plain  from  the 
three  following  propositions. 


PROP.  I.     PROBLEM. 


Three  magnitudes.  A,  B,  C  being  given,  to  find  a  fourth  such, 
that  every  three  shall  be  greater  than  the  remaining  one. 
Let  D  be  the  fourth  :  therefore  D  must  be  less  than  A,  B,  C 


344  NOTES. 

Book  XI.  together  :  of  the  three  A,  B,  C,  let  A  be  that  which  is  not  lesa 
"•— '''■■"^"'««'  tiiaii  eituer  of  the  two  B  and  C  :  and  first  let  B  and  C  together 
be  not  less  than  A  :  therefore  B,  C,  D  together  are  greater  than 
A  ;  and  because  A  is  not  less  than  B  ;  A,  C,  D  together  are 
gi'eater  than  B  :  in  the  like  manner  A,  B,  D  together  are  greater 
than  C  :  wherefore  in  the  case  in  which  B  and  C  together  are 
not  less  than  A,  any  magnitude  D  which  is  less  than  A,  B,  C 
together  will  answer  tiie  problem. 

But  if  B  and  C  together  be  less  than  A  ;  then,  because  it  is 
reqviired  tliat  B,  C,  D  together  be  greater  than  A,  from  each 
of  these  taking  away  B,  C,  the  remadning  one  D  must  be  greater 
than  the  excess  of  A  above  B  and  C  :  take  therefore  any  mag- 
nitude D  whici^.  is  less  than  A,  B,  C  together,  but  greater  than 
the  excess  of  A  above  B  and  C :  than  B,  C,  D  together  are  great- 
er than  A  ;  and  because  A  is  greater  than  eitner  B  or  C  much 
more  will  A  and  D,  together  with  either  of  the  two  t",  C  be 
greater  than  the  other  :  and,  by  the  construction,  A,  B,  C  are 
together  greater  than  D. 

CoR.  If  besides  it  be  required,  that  A  and  B  together  shall 
not  be  less  than  C  and  D  together ;  the  excess  of  A  and  i;  toge- 
ther above  C  must  not  be  less  than  D,  that  is,  D  must  not  be 
greater  than  that  excess. 


PROP.  II.    PROBLEM. 


Four  magnitudes  A,  E,  C,  D  being  given,  of  which  A  and 
B  together  are  not  less  than  C  and  D  together,  and  such  that 
any  three  of  them  whatever  are  greater  than  the  fourth  ;  it  is 
required  to  find  a  fifth  magnitude  E  such,  that  any  two  of  the 
three  A,  B,  E  shall  be  greater  than  the  third,  and  also  that  any 
two  of  the  three  C,  D,  E  shall  be  greater  than  the  third.  Let 
A  be  not  less  than  B:  and  C  not  less  than  D. 

First,  Let  the  excess  of  C  above  D  be  not  less  than  the  excess 
of  A  above  B  :  it  is  plain  that  a  magnitude  E  can  be  taken 
which  is  less  than  the  sum  of  C  and  D,  but  greater  than  the 
excess  of  C  above  D  ;  let  it  be  taken  :  then  E  is  greater  like- 
wise than  the  excess  of  A  above  B ;  wherefore  E  and  B  together 
are  greater  than  A ;  and  A  is  not  less  than  B  :  therefore  A  and 
E  together  are  greater  than  P. :  and,  by  the  hypothesis,  A  and 
B  together  are  not  less  than  C  and  D  together,  and  C  and  D 
together  are  greater  than  E ;  therefore  likewise  A  and  B  arc 
greater  than  E. 


NOTES.  345 

But  let  the  excess  of  A  above  B  be  greater  than  the  excess  of  Book  XI. 
C  above  D  :  and  because,  by  the  hypotnesis,  tiie  three  B,  C,  ^— ''^''^^w 
D  are  together  greater  than  the  fourth  A  ;  C  and  D  together 
are  greater  than  the  excess  of  A  above  B  :  therefore  a  magni- 
tude may  be  taken  whica  is  less  than  C  and  D  together,  but 
greater  than  the  excess  of  A  above  B.  Let  this  magnitude  be 
E  ;  and  because  E  is  greater  than  the  excess  of  A  above  B,  B 
together  with  E  is  greater  than  A  :  and,  as  in  the  preceding 
case,  it  may  be  shown  that  A  together  with  E  is  greater  than  B, 
and  that  A  together  with  B  is  greater  than  E  :  thei-efore,  in 
each  of  the  cases,  it  has  been  shown  that  any  two  of  the  three 
A,  B,  E  are  greater  than  the  third. 

And  because  in  each  of  the  cases  E  is  greater  than  the  excess 
of  C  above  D,  E  together  with  D  is  greater  than  C  ;  and,  by 
the  hypothesis,  C  is  not  less  than  D  ;  therefore  E  together  with 
C  is  greater  than  D  ;  and,  by  the  construction,  C  and  D  toge- 
ther are  greater  than  E  :  therefore  any  two  of  the  three,  C,  D, 
E  are  e;reater  than  the  third. 


PROP.  III.     THEOREM. 

There  may  be  innumerable  solid  angles  all  unequal  to  one 
another,  each  of  which  is  contdned  by  the  same  four  plane  an- 
gles, placed  in  the  same  order. 

Take  three  plane  angles.  A,  B,  C,  of  which  A  is  not  less 
than  either  of  the  other  two,  and  such,  that  A  and  B  toge- 
ther are  less  than  two  right  angles :  and  by  problem  1  and 
its  corollary,  find  a  fourth  angle  D  such,  that  any  three  what- 
ever of  the  angles  A,  B,  C,  D  be  greater  than  the  remaining 
angle,  and  such,  that  A  and  B  together  be  not  less  than  C 
and  D  together,:  and  by  problem  2,  find  a  fifth  angle  E  such 
that  any  two  of  the  angles  A,  B,  E  be  greater  than  the  third, 


-«nd  also  that  any  two  of  the  angles  C,  D,  E  be  greater  than 

2    X 


546 


NOTES. 


Book  XI.  the  third  :  and  because  A  and  B  together  are  less  than  two 
^^'^f^*^  right  angles,  the  double  of  A  and  B  together  is  less  than  four 
right  angles  :  but  A  and  B  together  are  greater  than  the  angle 
E  ;  wherefore  the  double  of  A,  B  together  is  greater  than 
the  three  angles  A,  B,  E  together,  which  three  are  conse- 
quently less  than  four  right  angles  ;  and  every  two  of  the 
same  angles  A,  B,  E  are  greater  tliun  the  third  ;  therefore, 
by  prop.  23,  1 ),  a  solid  angle  may  be  made  contained  by  three 
plane  angles  equal  to  the  angles  A,  B,  E,  each  to  each.  Let 
this  be  the  angle  F  contained  by  the  three  plane  angles  GFH, 
HFK,  GFK  wiiicn  are  equal  to  the  angles  A,  B,  E,  each  to 
each  :  and  because  the  angles  C,  D  together  are  not  greater 
than  the  angles  A,  B  together,  therefore  the  angles  C,  D,  E 
are  not  greater  than  the  angles  A,  B,  E :  but  these  last  three  are 
less  than  four  right  angles,  as  has  been  demonstrated  :  where- 
fore also  the  angles  C,  D,  E  are  together  less  than  four  right 
angles,  and  every  two  of  them  are  greater  than  the  third;  there- 
fore a  solid  angle  may  be  made  which  shall  be  contained  by  three 
a  23.  11.    plane  angles  equal  to  the  angles  C,  D,  E,  each  to  each  a:  and 


by  prop.  26,  1 1,  at  the  point  F  in  the  straight  line  FG  a  solid 
angle  may  be  made  equal  to  that  which  is  contained  by  the 
three  plane  angles  that  are  equal  to  the  angles  C,  D,  E  :  let 
this  be  made,  and  let  the  angle  GFK,  which  is  equal  to  E,  be 
one  of  the  three;  and  let  KFL,  GFL  be  the  other  two  which 
are  equal  to  the  angles,  C,  D,  each  to  each.  Thus  there  is  a 
solid  angle  constituted  at  the  point  F  contained  by  the  four  plane 
angles  GFH,  HFK,  KFL,  GFL  v/hich  are  equal  to  the  angles 
A,  B,  C,  D,  each  to  each. 

Again,  Find  another  angle  M  such,  that  every  two  of  the 
three  angles  A,  B,  M  l)e  greater  than  the  third,  and  also 
every  tv/o  of  the  three  C,  D,  M  be  greater  than  the  third : 


n. 


NOTES.  347 

and,  as  in  the  preceding    part,  it    may  be  demonsti-ated  that  Book  XI. 
the  three  A,  L-,  M  are  less  Vi^^v^**./ 

than  four  right  angles,  as  also 
that  the  three  C,  D,  M  are 
less  than  four  right  angles. 
Make  therefore  a  a  solid  angle 
at  N  contained  by  the  three 
plane  angles  ONP,  PNQ, 
ONQ,  which  are  equal  to  A, 
B,  M,  each  to  eacli :  and  by 
prop.    26,    11,  make   at  the 

point  N  in  the  straight  line  ON  a  solid  angle  contained  by  three 
plane  angles  of  which  one  is  the  angle  ONQ  equal  to  M,  and 
the  other  two  are  the  angles  QNR,  ONR  wiich  are  equal  to 
the  angles  C,  D,  each  to  each.  Thus,  at  the  point  N,  there 
is  a  solid  angle  contained  by  the  four  plane  angles  ONP,  PNQ, 
QNR,  ONR  which  are  equal  to  the  angles  A,  B,  C,  D,  each 
to  each.  And  that  the  two  solid  angles  at  the  points  F,  N, 
each  of  which  is  contained  by  the  above-named  fom-  plane  an- 
gles, are  not  equal  to  one  another,  or  that  they  cannot  coin- 
cide, will  be  plain  by  considering  that  the  angles  GFK,  ONQ; 
that  is,  the  angles  E,  M,  are  unequal  by  the  construction  ;  and 
therefore  the  straight  lines  GF,  i-K  cannot  coincide  with  ON, 
NQ,  nor  consequently  can  the  solid  angles,  which  therefore  are 
unequal. 

And  because  from  the  four  plane  angles  A,  B,  C,  D,  there 
can  be  found  innumtrable  other  angles  that  will  serve  the  same 
purpose  with  the  angles  E  and  M ;  it  is  plain  that  innumerable 
other  solid  angles  may  be  constituted  which  are  each  con.ained 
by  the  same  four  plane  angles,  and  all  of  them  unequal  to  one 
another.     Q.  E.  D. 

And  from  this  it  appears  that  Clavius  and  other  authors  are 
mistaken,  who  assert  that  those  solid  angles  are  eqvial  which  are 
contained  by  the  same  number  of  plane  angles  that  are  equal  to 
one  another,  each  to  each.  Also  it  is  plain  that  the  26th  prop, 
of  book  11,  is  by  no  means  sufficiently  demonstrated,  because 
the  equality  of  two  solid  angles,  whereof  each  is  contained  by 
three  plane  angles  which  are  equal  to  one  another,  each  to  each, 
15  only  assumed,  and  not  demonstrated. 


NOTES. 


PROP.  I.    B.  XI. 


The  words  at  the  end  of  this,  "  for  a  straight  line  cannot 
"  meet  a  straight  line  in  more  than  one  point,"  are  left  out,  as 
an  addition  by  some  unskilful  hand ;  for  tnis  is  to  be  demon- 
strated, not  assumed. 

Mr.  Tnomas  Simpson,  in  his  notes  at  the  end  of  the  2d  edition 
of  his  Elements  of  Geometry,  p.  252,  after  repeating  the  words 
of  this  note,  adds,  "  Noav,  can  it  possibly  show  any  want  of 
"  skill  in  an  editor  (he  means  Euciid  or  Theon)  to  refer  to  an 
"  axiom  which  Euclid  himself  hath  laid  down,  book  1,  No.  14, 
*'  (he  means  Barrow's  Euclid,  for  it  is  the  10th  in  the  Greek), 
"  and  not  to  have  demonstrated,  what  no  man  can  demonstrate  ?" 
But  all  that  in  tiiis  case  can  follow  from  that  axiom  is,  that,  if 
two  straight  lines  couid  meet  each  other  in  two  points,  the  parts 
of  them  betwixt  these  points  must  coincide,  and  so  they  would 
have  a  segment  betwixt  these  points  common  to  both.  Now,  as 
it  has  not  been  shown  in  Euciid,  that  they  cannot  have  a  com- 
mon segment,  this  does  not  prove  that  they  cannot  meet  in  two 
points  from  which  their  not  having  a  common  segment,  is  de- 
duced in  the  Greek  edition  :  but,  on  the  contrary,  because  they 
cannot  have  a  common  segment,  as  is  shown  in  cor.  of  1  Ith 
prop,  book  1,  of  4to  edition,  it  follows  plainly  that  they  cannot 
meet  in  two  points,  which  the  remarker  says  no  man  can  de- 
monstrate. 

Mr.  Simpson,  in  the  same  notes,  p.  265,  justly  observes,  that 
in  the  corollary  of  prop.  1  l,book  l,4to.  edition,  the  straight  lines 
AB,  BD,  BC,  are  supposed  to  be  all  in  the  same  plane,  which 
cannot  be  assumed  in  1st  prop,  book  11.  This,  soon  after  the 
4to.  edition  was  published,  I  observed  and  corrected  as  it  is  now 
in  this  edition  :  he  is  mistaken  in  thinking  the  10th  axiom  he 
mentions  here  to  be  Euclid's  ;  it  is  none  of  Euclid's  but  is  the 
10th  in  Dr.  Barrow's  edition,  who  had  it  from  Herigon's  Cur- 
sus,  vol.  1,  and  in  place  of  it  the  "corollary  of  10th  prop,  book  1, 
xyas  added. 


PROP.  II.     B.  XI. 

This  proposition  seems  to  have  been  changed  and  vitiated  by 
some  editor  :  for  all  the  figures  defipcd  in  the  first  book  of  the 
Elements,  arxl  among  them  triangles,  are,  by  the  hypothesis, 
plane  figures  ;  that  is,  such  as  are  described  in  a  plane ;  Avhere- 
fore  the  second  part  of  the  enunciation  needs  no  demonstration. 
Besides,  a  convex  superficies  may  be  terminated  by  three  straight 


NOTES.  349 

lines  meeting  one  another;  the  thing  that  should  have  been  de-Book  XI. 
monstrated  is,  that  two,  or  three  straight  iincs,  that  meet  one  ^•^'^^~^*'^ 
another  are  in  one  plane.     And  as  this  is  not  sufiiciently  done, 
the  enunciation  and  demonsti'ation  are  changed  into  those  now 
put  into  the  text. 

PROP.  III.     B.  XI. 

In  this  proposition  the  following  words  near  to  the  end  of  it 
are  left  out,  viz.  *'  therefore  DEB,  DFB  are  not  straight  lines  ; 
"  in  the  like  manner  it  may  be  demonstrated  that  there  can  be 
"  no  other  straight  line  between  the  points  D,  B :"  because  from 
this,  that  two  lines  include  a  space,  it  only  foilovv's  that  one  of 
them  is  not  a  straigiit  line  :  and  the  force  of  the  argument  lies 
in  t^iis,  viz.  if  the  common  section  of  the  planes  be  not  a  straight 
line,  then  tvvo  straight  lines  could  not  include  a  space,  which  is 
absurd  ;  therefore  the  common  section  is  a  straight  line. 

PROP.  IV.     B.  XI. 

The  words  "  and  the  triangle  AED  to  the  triangle  EEC"  are 
omitted,  because  the  Avhole  conclusion  of  the  4th  prop,  book  1. 
has  been  so  often  repeated  in  the  precedmg  books,  it  was  needless 
to  repeat  it  here. 

PROP.  V.     B.  XI. 

In  this,  near  to  the  end,  l5i-<Ts'Si<»,  ought  to  be  left  out  in  the 
Greek  text ;  and  the  word  "  plane"  is  rightly  left  out  in  the 
Oxford  edition  of  Commandine's  translation. 

PROP.  VII.     B.  XL 

This  proposition  has  been  put  into  this  book  by  some  un- 
skilful editor,  as  is  evident  from  this,  that  strdght  lines  which 
are  draAvn  from  one  point  to  another  in  a  plane,  are,  in  the 
preceding  books,  supposed  to  be  in  that  plane  :  and  if  they 
were  not  some  demonstrations  in  which  one  straight  line  is 
supposed  to  meet  another  would  not  be  conclusive,  because 
these  lines  would  not  meet  one  another  :  for  instance  in  prop. 
30,  book  1,  the  straight  lineGK  would  not  meet  EP',if  GK  were 
not  in  the  plane  in  which  are  the  parallels  AB,  CD,  and  in 
which,  by  hypothesis,  the  straight  line  EF  is  :  besides,  this 
7th  proposition  is  demonstrated  by  the  preceding  3d,  in  which 
the  very  thing  which  is  proposed  to  be  demonstrated  in  the  7th, 
is  twice  assumed,  viz.  that  the  straight  line  drawn  from  one 
point  to  another  in  a  plane,  is  in  that  plane  ;  and  the  same  thing 
is  assumed  in  the  preceding  6th  prop,  in  which  the  straight  line 


350  NOTES. 

Book  XI.  which  joins  the  points  13,  D  that  are  in  the  plane  to  which  AB, 
^-^~^^""*«*^  and  CD  are  ut  I'ight  tingles,  is  supposed  to  be  in  tuat  plane  : 
and  the  7th,  of  wixich  another  demonstration  is  givtn,  is  kept  in 
the  book,  merely  to  preserve  tne  number  ot  tue  propositions  : 
for  it  is  evident  from  the  7th  and  35th  definitions  ol  the  1st  book, 
though  it  had  not  been  in  the  elements. 

PROP.  VIII.     B.  XI. 

In  the  Greek,  and  in  Commandine's  and  Dr.  Gregory's  trans- 
lations, near  to  the  end  of  this  propo&iiion,  are  the  toilowing 
Avords:  "  but  DC  is  in  tiie  plane  tlirough  i.  A,  AD,"  insteaa  of 
Avliich,  in  tlie  Oxford  edition  of  Commanauie's  translation,  is 
rightly  put  "  but  DC  is  in  the  plane  through  rD,  DA:"  but 
all  the  editions  nave  the  following  words,  viz.  ''  because  Aii, 
"  liD  are  in  tiie  plane  through  L-D,  DA,  ana  DC  is  in  the  piane 
"  in  which  are  AB,  i>D,"  which  are  manifestly  corruptea,  or 
liave  been  added  to  the  text ;  for  there  was  not  the  least  neces- 
sity to  go  so  far  about  to  show  that  DC  is  in  the  s&me  piai:ie  in 
which  are  BD,  DA  because  it  immediately  follows  from  prop. 
7  preceding,  that  BD,  DA,  are  in  the  plane  in  winch  are  the 
parallels  Ab,  CD:  therefore,  instead  of  these  words,  there  ought 
only  to  be  "  because  all  three  are  in  the  plane  in  which  are  the 
parallels  AB,  CD." 

PROP.  XV.     B.  XI. 

After  the  words  "  and  because  EA  is  parallel  to  GH,"  the 
following  are  added,  "  for  each  of  them  is  parallel  to  DE,  and 
"  are  not  both  in  the  same  plane  with  it,"  as  being  manifestly 
forgotten  to  be  put  into  the  text. 

PROP.  XVI.     B.  XI. 

In  this,  near  to  the  end,  instead  of  the  Avords  "  but  straight 
"  lines  which  meet  neither  way"  ought  to  be  read,  "but  straight 
"  lines  in  the  same  plane  wiiich  produced  meet  neither  way  :" 
because,  though  in  citing  this  definition  in  prop.  27,  book  1,  it 
was  not  necessary  to  mention  the  words,  "  in  the  same  plane," 
all  the  straight  lines  in  the  books  preceding  this  being  in  the 
snme  plane  ;  yet  here  it  was  quite  necessary. 

PROP.  XX.     B.  XI. 

In  this,  near  the  beginning,  are  the  words,  "  But  if  not, 
"  let  f; AC  be  the  greater  :"  but  the  angle  BAG  may  happen  to 
he  equal  to  one  of  the  other  two  :  wherefoi'e  this  place  should 


NOTES.  351 

be  read  thus,  "  But  if  not,  let  the  angle  BAC  be  not  less  than  Book  XI. 
"  either  of  the  other  two,  but  greater  than  DAB."  v-o'^-^^v^ 

At  the  end  of  this  proposition  it  is  said,  "  in  the  same  man- 
"  ner  it  may  be  demonstrated,"  though  there  is  no  need  of  any 
demonstration;  because  the  angle  BAC  being  not  less  than 
either  of  the  other  two,  it  is  evident  that  BAC  together  with 
one  of  them  is  greater  than  the  other. 

PROP.  XXII.     B.  XL 

And  likewise  in  this,  near  the  beginning,  it  is  said,  "  But  if 
"  not,  let  the  angles  at  B,  E,  H  be  unequal,  and  let  the  angle 
"  at  B  be  greater  than  either  of  those  at  E,  H:"  which  Avords 
manifestly  show  this  place  to  be  vitiated,  because  the  angle  at 
B  may  be  equal  to  one  of  the  other  two.  They  ought  therefore 
to  be  read  thus,  "  But  if  not,  let  the  angles  at  B,  E,  H  be  une- 
"  qual,  and  let  the  angle  at  B  be  not  less  than  either  of  the 
"  other  two  at  E,  H:  therefore  the  straight  line  AC  is  not  less 
"  than  either  of  the  two  DF,  UK." 

PROP.  XXIII.     B.  XI. 

The  demonstration  of  this  is  made  something  shorter,  by 
not  repeating  in  the  third  case  the  things  which  were  demon- 
strated in  the  first ;  and  by  making  use  of  the  construction  which 
Campanus  has  given;  but  he  does  not  demonstrate  the  second 
and  third  cases;  the  construction  and  demonstration  of  the  third 
ease  are  made  a  little  more  simple  than  in  the  Oreek  text. 

PROP.  XXIV.     B.  XI. 

The  word  "  similar"  is  added  to  the  enunciation  of  this  pro- 
position, because  the  planes  containing  the  solids  wnich  are  to 
be  demonstrated  to  be  equal  to  one  another,  in  the  25th  propo- 
sition, ought  to  be  similar  and  equal,  that  the  equality  of  the 
solids  may  be  inferred  from  prop.  C,  of  this  book :  and  in  the 
Oxford  edition  of  Commandine's  translation,  a  corollary  is  add- 
ed to  prop.  24,  to  show  that  the  parallelograms  mentioned  in 
this  proposition  are  similar,  that  the  equality  of  the  solids  in 
prop.  25,  may  be  deduced  from  the  10th  def.  of  book  1 1. 

PROP.  XXV.  and  XXVI.     B.  XI. 

In  the  25th  prop,  solid  figures,  which  are  contained  by  the 
same  number  of  similar  and  equal  plane  figures,  are  supposed 


352  NOTES. 

Rook  XI.  to  be  equul  to  one  another.  And  it  seems  that  Theon,  or  some 
"W^v-^XBi--  other  editor,  that  he  might  save  himseif  the  trouble  of  demon- 
strating the  soiid  figures  mentioned  in  this  proposition  to  be 
equal  to  one  another,  has  inserted  the  10th  clef,  of  tl^is  book, 
to  serve  instead  of  a  demonstration ;  which  was  very  ignorantly 
done. 

Likewise  in  the  26th  prop,  two  solid  angles  are  supposed  to 
be  equal :  if  each  of  them  be  contained  by  three  plane  angles 
Avhich  are  equal  to  one  another,  each  to  each.  And  it  is  strange 
enough,  that  none  of  the  commentators  on  Euclid  have,  as  far 
as  I  know,  perceived  that  sometning  is  vx^ancing  in  the  demon- 
strations of  these  two  propositions.  Clavius,  indeed,  in  a  note 
upon  the  1 1th  def.  of  tiiis  book,  affirms,  that  it  is  evident  that 
those  solid  angles  are  even  which  are  contained  by  the  same 
number  of  plane  angles,  equal  to  one  another,  each  to  each, 
because  they  will  coincide,  if  they  be  conceived  to  be  placed 
within  one  another  ;  but  this  is  sa.id  without  any  proof,  nor  is  it 
ahvays  true,  except  when  the  solid  angles  are  contained  by  three 
plane  angles  only,  Avhich  are  equal  to  one  another,  each  to 
each :  and  in  this  case  the  proposition  is  the  same  with  this, 
that  two  spherical  triangles  that  are  equilateral  to  one  another, 
are  also  equiangular  to  one  another,  and  can  coincide ;  which 
ought  not  to  be  granted  without  a  demonstration.  Euclid  does 
not  assume  this  in  the  case  of  rectilineal  triangles,  but  demon- 
strates, in  prop.  8,  book  1,  that  triangles  which  are  equilateral 
to  one  another  are  also  equiangvilar  to  one  another  ;  and  from 
this  their  total  equality  appears  by  prop.  4,  book  1 .  And  Me- 
nelaus,  in  the  4th  prop-,  of  his  1st  book  of  spherics,  explicitly 
demonstrates,  that  spherical  triemgles  which  are  mutually  equi- 
lateral, are  also  equiangular  to  one  another  ;  from  which  it  is 
easy  to  show  that  they  must  coincide,  pro\iding  they  have  their 
sides  disposed  in  the  same  order  and  situation. 

To  supply  these  defects,  it  was  necessary  to  add  the  thi'ee 
propositions  marked  A,  B,  C  to  this  book.  I'or  the  25th,  26th, 
and  28th  propositions  of  it,  and  consequently  eight  others,  viz. 
the  27th,  31st,  32d,  33d,  34th,  36th,  37th,  and  40th  of  the  same, 
which  depend  upon  them,  h-ave  hitherto  stood  upon  an  infirm 
foundation;  as  also,  the  8th,  12th,  cor.  of  17th  and '18th  of 
j^  12th  book,  v/hich  depend  upon  the  9th  definition.  For  it  has 
been  shov/n  in  the  notes  on  def.  10,  of  this  book,  that  solid 
figures  which  are  contained  by  the  same  number  of  similar  and 
equal  plane  figvu'cs,  as  also  solid  angles  that  are  cont.dned  by 
the  same  number  of  equal  plane  angles,  are  not  always  equal 
to  one  another. 


NOTES.  353 

It  is  to  be  observed  that  Tacquet,  in  his  Euclid,  defines  equal  Book  XI. 
solid  angles  to  be  such,  "  as  being  put  within  one  another  do  ^-^"v^>-.' 
"  coincide  :"  but  this  is  an  axiom,  not  a  definition  ;  for  it  is  true 
of  all  magnitudes  whatever.  He  made  this  useless  definition, 
that  by  it  he  might  demonstrate  the  36th  prop,  of  tnis  book, 
without  the  help  of  the  35th  of  the  same  :  concerning  wnich 
demonstration,  see  the  note  upon  prop.  36. 

PROP.  XXVIII.     B.  XL 

In  this  it  ought  to  have  been  demonsti-ated,  not  assumed, 
that  the  diagonals  are  in  one  plane.  Clavius  had  supplied  l.Js 
defect. 

PROP.  XXIX.     B.  XL 

There  are  three  cases  of  this  proposition  ;  the  first  is,  when 
the  two  parallelograms  opposite  to  the  base  AB  have  a  side 
common  to  both  ;  the  second  is,  when  these  parallelograms  are 
separated  from  one  another ,  and  the  thii'd,  when  there  is  a  part 
of  them  common  to  both  ;  and  to  this  last  only,  the  demonstra- 
tion that  has  hitherto  been  in  the  Elements  does  agree.  The 
first  case  is  immediately  deduced  from  the  preceding  28th 
prop,  which  seems  for  this  purpose  to  have  been  premised  to 
this  29th,  for  it  is  necessary  to  none  but  to  it,  and  to  the  40th 
of  this  book,  as  we  now  have  it,  to  which  last  it  would,  without 
doubt,  have  been  premised,  if  Euclid  had  not  made  use  of  it  in 
the  29th;  but  some  unskilful  editor  has  taken  it  away  from  the 
Elements,  and  has  mutilated  Euclid's  demonstration  of  the 
other  two  cases,  which  is  now  restored,  and  serves  for  both  at 
once. 

PROP.  XXX.     B.  XL 

In  the  demonstration  of  this,  the  opposite  planes  of  the  solid 
CP,  in  the  figure  in  this  edition,  that  is,  of  the  solid  CO  in 
Commandine's  figure,  are  not  proved  to  be  parallel ;  which  it  is 
proper  to  do  for  the  sake  of  learners. 

PROP.  XXXI.     B.  XL 

There  are  two  cases  of  this  proposition  ;  the  first  is,  when 
the  insisting  straight  lines  are  at  right  angles  to  the  bases  ;  the 
other,  when  they  are  not :  the  first  case  is  divided  again  into 
two  others,  one  of  which  is,  when  the  bases  are  equiangular 
parallelograms ;  the   other  when  thev   are   not  equiangular : 

2  Y  ■ 


354  NOTES. 

Book  XI.  the  Greek  editor  makes  no  mention  of  the  first  of  these  two 
^"^'■"''">>-^  last  cases,  but  has  inserted  the  demonstration  of  it  as  a  part  of 
that  of  the  other :  and  tiierefore  should  have  taken  notice  of  it 
in  a  corollary ;  but  we  thought  it  better  to  give  these  two  cases 
separately  :  the  demonstration  also  is  made  something  shorter 
by  following  the  way  Euclid  has  made  use  of  in  prop.  14,  book  6. 
Besides,  in  the  demonstration  of  the  case  in  which  the  insisting 
straight  lines  are  not  at  right  angles  to  the  bases,  the  editor  does 
not  prove  that  the  solids  described  in  the  construction  are  paral- 
lelopipeds,  which  it  is  not  to  be  thought  that  Euclid  neglected : 
also  the  words  "  of  which  the  insisting  straight  lines  are  not  in 
"  the  same  straight  lines,"  have  been  added  by  some  unskilful 
hand ;  for  they  may  be  in  the  same  straight  lines. 


PROP.  XXXII.     B.  XI. 

The  editor  has  forgot  to  order  the  parallelogram  FH  to  be 
applied  in  tlie  angle  FGH  equal  to  the  angle  LCG,  which  is 
necessary.     Clavius  has  supplied  this. 

Also,  in  the  construction,  it  is  required  to  complete  the  so- 
lid of  which  the  base  is  FH,  and  altitude  the  same  with  that 
of  the  solid  CD  :  but  this  does  not  determine  the  solid  to  be 
completed,  since  there  may  be  innumerable  solids  upon  the 
same  base,  and  of  the  same  altitude  :  it  ought  therefore  to  be 
said  "  complete  the  solid  of  which  the  base  is  FH,  and  one  of 
"  its  insisting  straight  lines  is  FD  ;"  the  same  correction  must 
be  made  in  the  following  proposition  33. 


PROP.  D.     B.  XI. 

It  is  very  probable  that  Euclid  gave  this  proposition  a  place 
in  the  Elements,  since  he  gave  the  like  proposition  concerning 
equiangular  parallelograms  in  the  23d  B.  6. 


PROP.  XXXIV.     B.  XI. 

In  this  the  words,  uv  «/  itpis-raa-t  ax.  utrtv  Izri  tmv  civruv  ivhiav, 
"  of  which  the  insisting  straight  lines  are  not  in  the  same 
"  straight  lines,"  arc  thrice  repeated ;  but  these  words  ought 
either  to  be  left  out,  as  they  are  by  Clavius,  or,  in  place  of  them, 
ought  to  be  put,  "  whether  the  insisting  straight  lines  be,  or  be 
"  not,  in  the  same  straight  lines:"  for  the  other  case  is  with- 
out any  reason  excluded ;  also  the  words,  *v  roi  -xln,  of  which 


NOTES.  35b 

"  the  altitudes,"  are  twice  put  for  uv  ^ai  'iipic-Tucrcit,  "  of  which  Book  XI. 
"  the  insisting  straight  lines;"  which  is  a  plain  mistake:  for '^■-^"^■^^^^ 
the  altitude  is  always  at  right  angles  to  the  base. 

PROP.  XXXV.     B.  XI. 

The  angles  ABH,  DEM.are  demonstrated  to  be  right  angles 
in  a  shorter  way  than  in  the  Greek  ;  and  in  the  same  Avay  ACH, 
DEM  may  be  demonstrated  to  be  right  angles  :  also  the  i-epe- 
tition  of  the  same  demonstration,  which  begins  with  "  in  the 
"  same  manner,"  is  left  out,  as  it  was  probably  added  to  the 
text  by  some  editor  ;  for  the  words,  "  in  like  manner  we  may 
"  demonstrate,"  are  not  inserted  except  when  the  demonstration 
is  not  given,  or  when  it  is  something  different  froin  the  other 
if  it  be  given,  as  in  prop.  26,  of  this  book.  Companus  has  not 
this  repetition. 

We  have  given  another  demonstration  of  the  corollary,  be- 
sides the  one  in  the  original,  by  help  of  which  the  following 
36th  prop,  may  be  demonstrated  without  the  35th. 

PROP.  XXXVI.     B.  XL 

Tacquet  in  his  Euclid  demonstrates  this  proposition  without 
the  help  of  the  35th  ;  but  it  is  plain,  that  the  solids  mentioned 
in  the  Greek  text  in  the  enunciation  of  the  proposition  as  equi- 
angular, are  such  that  their  solid  angles  are  contained  by  thi-ee 
plane  angles  equal  to  one  another,  each  to  each  ;  as  is  evident 
from  the  construction.  Now  Tacquet  does  not  demonstrate, 
but  assumes  these  solid  angles  to  be  equal  to  one  another  ;  for 
he  supposes  the  solids  to  be  already  made,  and  does  not  give 
the  construction  by  Avhich  they  are  made  :  but,  by  the  second 
demonstration  of  the  preceding  corollary,  his  demonstration  is 
rendered  legitimate  likewise  in  the  case  where  the  solids  are 
constructed  as  in  the  text. 

PROP.  XXXVII.     B.   XI. 

In  this  it  is  assumed  that  the  ratios  which  are  triplicate  of 
those  ratios  which  are  the  same  with  one  another,  are  likewise  . 
the  same  with  one  another  ;  and  that  those  ratios  are  the  same 
with  one  another,  of  which  the  triplicate  ratios  are  the  same 
with  one  another  ;  but  this  ought  not  to  be  granted  without  a 
demonstration  ;  nor  did  Euclid  assume  the  first  and  easiest  of 
these  two  pi'opositions,  but  demonstrated  it  in  the  case  of  dupli- 
cate ratios,  in  the  22d  prop,  book  6.  On  this  account,  another 
demonstration  is  given  of  this  proposition  like  to  that  which 
Euclid  gives  in  prop.  22,  book  6,  as  Clavius  has  done. 


356  NOTES. 

Book  XI. 

v^-v-^^  PROP.  XXXVIII.     B.  XI. 

< 

When  it  is  required  to  draw  a  perpendicular  from  a  point  in 
one  plane  wldch  is  at  right  angles  to  another  plane,  unto  tliis 
last  piane,  it  is  done  by  drawing  a  perpcntlicular  from  the  point 
*  to  tiie  common  section  of  the  planes  ;  for  this  perpendicular 

will  be  perpendicular  to  the  plane,  by  def.  4,  of  tlus  book  : 
and  it  would  be  foolish  in  this  case  to  do  it  by  the  1 1th  prop,  of 
•^y  -ic,  In  the   same:   but   Euclid  a,  ApoUonius,   and  other   geometers, 
otl  e   (Tdi-  when  they  have  occasion  for  this  problem,  direct  a  perpendicu- 
tions.  lar  to  be  drawn  from  the  point  to  the  plane,  and  conclude  that 

it  will  fail  upon  the  common  section  of  the  planes,  because  this 
is  the  very  same  thing  as  if  they  had  made  use  of  the  construc- 
tion above  mentioned,  and  then  concluded  that  the  straight  line 
must  be  perpendicular  to  the  plane  ;  but  is  expressed  in  fewer 
words.  Some  editor,  not  perceiving  tiiis,  thought  it  was  ne- 
cessary to  add  tiiis  proposition,  wiiich  can  never  be  of  any  use 
to  the  1 1th  book,  and  its  being  near  to  the  end  among  proposi- 
tions with  which  it  has  no  connexion,  is  a  mark  of  its  having 
been  added  to  the  text. 

PROP.  XXXIX.     B.  XI. 

In  this  it  is  supposed,  that  the  straight  lines  which  bisect  the 
sides  of  the  opposite  planes,  are  in  one  plane,  which  ought  to 
have  been  demonstrated  ;  as  is  now  done. 


BOOK  XII. 

Book  XII.  THE  learned  Mr.  Moore,  professor  of  Greek  in  the  Univer- 
^^r-s^->^  sity  of  Glasgow,  observed  to  me,  that  it  plainly  appears  from 
Archimedes's  epistle  to  Dositheus,  prefixed  to  his  books  of  the 
Sphere  and  Cylinder,  which  epistle  he  has  restored  from  anci- 
ent manuscripts,  that  Eudoxus  was  the  author  of  the  chief  pro- 
positions in  this  12th  book. 

PROP.  II.     B  XII. 

At  the  beginning  of  this  it  is  said,  "  if  it  be  not  so,  the  square 
"  of  BD  shall  be  to  the  square  of  FH,  as  the  circle  AKCD  is 
"  to  some  space  either  less  than  the  circle  EFGH,  or  greater 
"  than  it."  And  the  like  is  to  be  found  near  to  the  end  of  this 
proposition,  as  also  in  prop.  5,   11,   12,  18,  of  this  book  :  con- 


NOTES.  357 

cerning  which,  it  is  to  be  observed,  that,  in  the  demonstration  Book  XII. 
of  theorems,  it  is  sulhcient,  in  tiiis  and  the  iik;_  cses,  that  a  s^'v^^i' 
thing-  made  use  of  in  the  I'easoning  can  possibly  exist,  pro\i- 
ding  tuis  be  evident,  though  it  cannot  be  exhibited  or  fovmd  by 
a  geometrical  coiistruction :  so,  in  tiis  place,  it  is  assumed,  that 
there  may  be  a  fouitn  proportional  to  these  three  mui^nitudes, 
viz.  the  squares  of  D,  t  H,  and  tlie  circle  A  CD  ;  because 
it  is  evident  that  tilers;  is  some  square  equa;  to  the  circle  ABCD 
though  it  cannot  be  found  geometrically  ;  and  to  the  three  recti- 
lineal figures,  viz.  the  squares  of  BD,  FH,  and  the  square 
which  is  equal  to  the  circle  ABCD,  there  is  a  fourth  square 
proportional  ;  because  to  the  three  stridght  lines  which  are 
their  sides,  there  is  a  fourth  straight  line  proportional  a,  and  a  12.  6. 
this  fourth  square,  or  a  space  equal  to  it,  is  the  space  which 
in  this  proposition  is  denoted  by  the  letter  S  :  and  tiie  like  is  to 
be  understood  in  tiie  other  places  above  cited  :  and  it  is  proba- 
ble tnat  this  has  been  shown  by  Euclid,  but  left  out  by  some 
editor  ;  for  the  lemma  which  some  unskilful  hand  has  added  to 
this  proposition  explains  nothing  of  it. 

PROP.  III.     B.  XII. 

In  the  Greek  text  and  the  translations,  it  is  said,  "  and 
"  because  the  two  straight  lines  BA,  A-  which  meet  one  ano- 
"  ther,"  &:c.  here  the  angles  BAC,  KHL  are  demonstrated 
to  be  equal  to  one  another  by  10th  prop.  B.  11,  which  had 
been  done  before  :  because  the  triangle  EAG  was  proved  to  be 
similar  to  the  triangle  KHL  :  this  repetition  is  left  out,  and 
the  triangles  BAC,  KHL,  are  proved  to  be  similar  in  a  shorter 
way  by  prop.  21,  B.  6. 

PROP.  IV.     B.  XII. 

A  few  things  in  this  are  moi'e  fully  explained  than  in  the 
Greek  text. 


PROP.  V.     B.  XII. 

In  this,  near  to  the  end,  are'  the  words,  »?  s^;t§05-5sv  ihx6>t, 
"  as  was  before  shown,"  and  the  same  are  found  again  m  the 
end  of  prop.  18,  of  this  book  :  but  the  demonstration  referred 
to,  except  it  be  the  useless  lemma  annexed  to  the  2d  prop,  is  no 
where  in  these  Elements,  and  has  been  perhaps  left  out  by  some 
editor  who  has  forgot  to  cancel  those  words  also. 


NOTES. 


PROP.  VI.     B.  XII. 


A  shorter  demonstration  is  sriven  of  this  ;  and  that  which 
is  in  the  Greek  text  may  be  made  sliorter  by  a  step  than  it  is  : 
for  the  author  of  it  makes  use  of  the  22d  prop,  of  B.  5,  twice  : 
whereas  once  would  have  served  his  purpose  ;  because  that 
proposition  extends  to  any  number  of  magnitudes  which  are 
proportionals  taking  two  and  two,  as  well  as  to  three  which  are 
proportional  to  other  three. 

COR.     PROP.   Vm.     B.   XII. 


a20.  6- 

bll.def. 

11. 

c4.  6. 


The  demonstration  of  this  is  imperfect,  because  it  is  not 
shown,  that  the  triangular  pyram.ids  into  which  those  upon 
multangular  bases  are  divided,  are  similar  to  one  another,  as 
ought  necessarily  to  have  been  done,  and  is  done  in  the  like 
case  in  prop.  12,  of  this  book.  The  full  demonstration  of  the 
corollary  is  as  follows  : 

Upon  the  polygonal  bases  ABCDE,  FGHKL,  let  there  be  si- 
milar and  similarly  situated  pyramids  which  have  the  points 
M,  N  for  their  vertices  :  the  pyramid  ABCDEM  has  to  the 
pyramid  FGHKLN  the  triplicate  ratio  of  that  which  the  side 
AB  has  to  the  homologous  side  FG. 

Let  the  polygons  be  divided  into  the  triangles  ABE,  EBC, 
ECD  ;  FGL,  LGH,  LHK,  which  are  similar  a  each  to  each  ; 
and  because  the  pyramids  are  similai',  thereforeb  the  triangle 
EAM  is  similar  to  the  triangles  LFN,  and  the  triangle  ABM 
to  FGN  :  wherefore  c  ME  is  to  EA,  as  NL  to  LP  ;  and  as  AE 


C     L 


A  B  F 

to  EB,  so  is  FL  to  LG,  because  the  triangles  EAB,  LFG  are 
similar  ;  therefore,  ex  jequali,  as  ME  to  EB,  so  is  NL  to  LG  : 


NOTES.  359 

in  like  manner  it  may  be  shown  that  EB  is  to  BM,  as  LG  to  Book  XII. 
GN  ;  therefore  again  ex  aquali^  as  EM  is  to  MB,  so  is  LN  to  ^-^^^^s-/ 
GN  :  wherefore  the  triangles  EMB,  LNG  having  their  sides 
proportionals  are  a  equiangular,  and  similar  to  one  another  :  a  5.  6. 
therefore  the  pyramids  which  have  the  triangles  EAB  LFG  for 
their  bases,  and  the  points  M,  N  for  their  veitices,  are  similar 
b  to  one  another,  for  their  solid  angles  are  c  equal,  and  theb  11.  clef, 
solids  themselves  are  contained  by  the  same  number  of  similar  bi- 
planes :   in  the  same  manner,  the  pyramid  EBCM  may   becb.ll. 
shown  to  be  similar  to  the  pyramia  LGHN,  and  the  pyramid 
ECDM  to  LHKN.   And  because  the  pyramids  EABM,  LFGN 
are  similar,  and  have  triangular  bases,  the  pyramid  EABM  has 
d  to  LFGN  the  triplicate  ratio  of  that  which  EB  has  to  the  ho-d  8.  12. 
mologous  side  LG.     And  in  the  same  inanner,  the  pyramid 
EBCM  has  to  the  pyramid  LGHN  the  triplicate  ratio  of  that 
which  EB  has  to  LG.     Therefore  as  the  pyramid  EABM  is  to 
the  pyramid  LFGN,  so  is  the  pyramid  EBCM  to  the  pyramid 
LGHN.     In  like  manner  as  the  pyramid  EHCM  is  to  LGHN, 
so  is  the  pyramid  ECDM  to  the  pyramid  LHKN.    And  as  one 
of  the  antecedents  is  to  one  of  the  consequents,  so  are  all  the 
antecedents  to  all  the  consequents  :  therefore  as  the  pyramid 
EABM  to   the   pyramid   LiGN   so  is  the    whole    pyramid 
ABCDEM  to  the  whole  pyramid  FGHKLN  :  and  the  pyramid 
EABM  has  to  the  pyramid  LtGN  the  triplicate  ratio  of  that 
which  AB  has  to  FG  ;  therefore  the  whole  pyramid  has  to  the 
whole  pyramid  the  triplicate  ratio  of  that  which  AB  has  to  the 
homologous  side  FG.     Q.  E.  D. 

PROP.  XL  and  XH.     B.  XH. 

The  order  of  the  letters  of  the  alphabet  is  not  observed  in  these 
two  propositions,  according  to  Euclid's  manner,  and  is  now 
restored  ;  by  which  means  the  first  part  of  prop.  12  may  be  de- 
monstrated in  the  same  words  with  the  first  part  of  prop.  II: 
on  this  account  the  demonstration  of  that  first  part  is  left  out, 
and  assumed  from  prop.  1 1 . 

PROP.  XH.     B.  XH. 

In  this  proposition  the  common  section  of  a  plane  parallel  to 
the  bases  of  a  cylinder,  with  the  cylinder  itself,  is  supposed  to  be 
a  circle,  and  it  was  thought  proper  briefly  to  demonstrate  it ; 
from  whence  it  is  sufficiently  manifest,  that  this  plane  divides 
the  cylinder  into  two  others  ;  and  the  same  thing  is  understood 
to  be  supplied  in  prop.  14. 


NOTES. 


PROP.  XV.     B.  XII. 

"  And  complete  the  cylinders  AX,  EO,"  both  the  enuncia- 
tion and  exposition  of  the  proposition  represent  the  cylinders  as 
well  as  the  cones,  as  already  described  :  wiierefore  the  reading 
ought  rather  to  be,  "  and  let  the  cones  be  ALC,  ENG  ;  and  the 
«  cylinders  AX,  EO." 

The  first  case  in  the  second  part  of  the  demonstration  is  want-- 
ing  ;  and  something  also  in  the  second  case  of  that  part,  before 
the  repetition  of  the  construction  is  mentioned  ;  which  are  now 
added. 


PROP.  XVII.     B.  XII. 

In  the  enunciation  of  this  proposition,   the   Greek  words 

Ui  Ts;v  Ltii?ovx  cpciipxv  c-neiov  TioXviO^ov  iyTpor^xi  jt4>iy«t/«v  t*j5  iXutrtJOVc? 

(7(pxte^cii  yMTccTYiv  i-pripdvuiiv  are  thus  translated  by  Ccmmandine 
and  others,  "  in  majore  solidum  polyhedrum  describere  quod 
"  minoris  sphxrs;  superficiem  non  tangat ;"  that  is,  "  to  de- 
"  scribe  in  the  greater  sphere  a  solid  polyhedron  which  shall 
"  not  meet  the  superficies  of  the  lesser  sphere  :"  whereby 
they  refer  the  words  Kara  ryiv  iTricpavuocy  to  these  next  to  th.em 
T>)«-  iKaaaovo?  a-tpui^ui.  I'ut  they  ought  by  no  means  to  be  tiius 
translated ;  for  the  solid  polyhedron  doth  not  only  meet  the 
superficies  of  the  lesser  sphere,  but  pervades  the  whole  of  that 
sphere  ;  therefore  the  foresaid  words  are  to  be  referred  to 
TO  e-Tsg5flv  7ro>.ii5?§«v,  and  ought  thvis  to  be  translated,  viz  to  describe 
in  the  greater  sphere  a  solid  polyhedron  Avhose  superficies  shall 
not  meet  the  lesser  sphere  ;  as  the  meaning  of  the  proposition 
necessarily  requires. 

The  demonstration  of  the  proposition  is  spoiled  and  mutilat- 
ed :  for  some  easy  things  are  very  explicitly  demonstrated, 
Avhile  others  not  so  obvious  are  not  sufficiently  explained  :  for 
example,  when  it  is  affirmed,  that  tl^iC  square  of  Kb  is  greater 
than  the  double  of  the  square  FZ,  in  the  first  demonstra- 
tion, and  that  the  angle  f'ZK  is  obtuse,  in  the  second  ;  both 
which  ought  to  have  been  demonstrated.  Besides,  in  the  first 
demonstration  it  is  said,  "draw  KC2  from  the  point  K  perpen- 
"  dicular  to  BD  ;"  whereas  it  ought  to  have  been  said,  "  join 
"  KV,"  and  it  should  have  deen  demonstrated  that  KA^  is 
perpendicular  to  liD  :  for  it  is  evident  from  the  figure  in  Her- 
vagius's  and  Gregory's  editions,  and  from  the  words  of  the 


NOTES.  561 

demonstration,  that  the  Greek  editor  did  not  perceive  that  Book  XII. 
the  perpendicular  drawn  from  the  point  K  to  the  straight  line  ^^^'v-^s-^ 
BD  must  necessarily  fall  upon  the  point  V,  for  in  the  figure  it 
is  made  to  fall  upon  the  point  Q  a  different  point  from  V, 
which  is  likewise  supposed  in  the  demonstration.  Comman- 
dine  seems  to  have  been  aware  of  this  ;  for  in  this  figure  he 
marks  one  and  the  same  point  with  the  two  letters  V,  12  ;  and 
before  Commandine,  the  learned  John  Dee,  in  the  commen- 
tary he  annexes  to  this  proposition  in  Henry  Billinsley's  trans- 
lation of  the  Elements,  printed  at  London,  anno  1570,  expressly 
takes  notice  of  this  error,  and  gives  a  demonstration  suited  to 
the  construction  in  the  Greek  text,  by  Avhich  he  shows  that  the 
perpendicular  drawn  from  the  point  K  to  BD,  must  necessarily 
fall  upon  the  point  V. 

Likewise  it  is  not  demonstrated  that  the  quadrilatei'al  figures 
SOFT,  TPRY  and  the  triangle  YRX  do  not  meet  the  lesser 
sphere,  as  was  necessary  to  have  been  done  :  only  Clavius,  as 
far  as  I  know,  has  observed  this,  and  demonstrated  it  by  a  lem- 
ma, which  is  now  premised  to  this  proposition,  something  alter- 
ed and  more  briefly  demonstrated. 

In  the  corollary  of  this  proposition,  it  is  supposed  that  a  solid 
polyhedron  is  described  in  the  other  sphere  similar  to  that  which 
is  described  in  the  sphere  BCDE  ;  but,  as  the  construction  by 
which  this  may  be  done  is  not  given,  it  was  thought  proper  to 
give  it,  and  to  demonstrate,  that  the  pyramids  in  it  are  similar 
to  those  of  the  same  order  in  the  solid  polyhedron  described  ii) 
the  sphere  BCDE. 

From  the  preceding  notes,  it  is  sufficiently  eAident  how  much 
the  Elements  of  Euclid,  who  was  a  most  accurate  geometer, 
have  been  vitiated  and  mutilated  by  ignorant  editors.  The 
opinion  which  the  greatest  part  of  learned  men  have  entertained 
concerning  the  present  Greek  edition,  viz.  that  it  is  very  little  or 
nothing  different  from  the  genuine  work  of  Euclid,  has  without 
doubt  deceived  them,  and  made  them  less  attentive  and  accurate 
in  examining  that  edition ;  whereby  several  errors,  some  of  them  ^ 
gross  enough,  have  escaped  their  notice  from  the  age  in  which 
Tlieon  lived  to  this  time.  Upon  which  account  there  is  some 
ground  to  hope,  that  the  pains  we  have  taken  in  correcting  those 
errors,  and  freeing  the  Elements,  as  far  as  we  could,  from  ble- 
mishes, will  not  be  unacceptable  to  good  judges,  who  can  discern 
when  demonstrations  are  legitimate,  and  when  they  are  not. 

The  objections  which,  since  the  first  edition,  have  been  made 
against  some  things  in  the  notes,  especially  against  the  doctrine 
of  proportionals,  have  either  been  fully  answered  in  Dr.  Bar- 
row's Lect.  Mathemat.  and  in  these  notes,  or  are  such,  except 

2  Z 


362  NOTES. 

Book  XII.  one  which  has  been  taken  notice  of  in  the  note  on  Prop.  1.  Book 
^-^'"-'''''"^^  1 1 .  as  show  that  the  person  who  made  them  has  not  sufficiently 
considered  the  things  against  which  they  are  brought  :  so  that 
it  is  not  necessary  to  make  any  further  answer  to  these  objections 
and  others  like  them  against  Euclid's  definition  of  proportionals; 
of  which  definition  Dr.  Barrow  justly  says,  in  page  297  of  the 
above  named  book,  that  "  Nisi  machinis  impulsa  validioribus 
"  seternum  persistet  inconcussa." 


FINIS. 


«BttcIitfS  ^aU. 


THIS  EDITION 


SEVERAL  ERRORS  ARE  CORRECTED, 


AND 


SOME  PROPOSITIONS  ADDED. 


BY  ROBERT  SIMPSOJ^T,  M.  D. 

EMERITUS    PROFESSOR    OF    MATHEMATICS    IN    THE 
UNIVERSITY    OF    GLASGOW. 


PHILADELPHIA 


PRINTED    BY    WM.    F.   m'LAUGHLIN,    NO.    28    NORTH 
SECOND    STREET. 

1806. 


PREFACE. 


EUCLID'S  DATA  is  the  first  in  order  of  tlie  books 
WTitten  by  the  ancient  geometers  to  facilitate  and  pro- 
mote the  method  of  resolution  or  analysis.  In  the 
general,  a  thing  is  said  to  be  given  which  is  either 
actually  exhibited,  or  can  be  found  out,  that  is,  which 
is  either  knoA\'n  by  h}'pothesis,  or  that  can  be  demon- 
strated to  be  known;  and  the  propositions  in  the 
book  of  Euclid's  Data  show  'what  things  can  be  found 
out  or  known  from  those  that  by  h}"pothesis  are  al- 
ready knoA\  n ;  so  that  in  the  analysis  or  investigation 
of  a  problem,  from  the  thiiigs  that  are  laid  down  to 
be  known  or  given,  by  the  help  of  these  propositions 
other  things  are  demonstrated  to  be  given,  and  from 
these,  other  things  are  again  sho^^'n  to  be  given,  and 
so  on,  until  that  which  was  proposed  to  be  fo-nid  out 
in  the  problem  is  demonstrated  to  be  given,  and  \\  hen 
this  is  done,  the  problem  is  solved,  and  its  composi- 
tion is  made  and  derived  from  the  compositions  of  the 
Data  v.  hich  were  made  use  of  in  the  analysis.  And 
thiws  the  Data  of  Euclid  are  of  the  most  general  and 
necessary  use  in  the  solution  of  problems  of  ever}- 
kind. 

Euclid  is  reckoned  to  be  the  author  of  the  Book 
of  the  Data,  both  by  the  ancient  and  modern  geome- 
ters ;  and  there  seems  to  be  no  doubt  of  his  having 
wri'ttn  a  book  on  this  subject,  but  which,  in  the 
course  of  so  niany  ages,  has  been  much  vitiated  by 


366  PREFACE. 

unskilful  editors  in  several  places,  both  in  the  order 
of  the  propositions,  and  in  the  definitions  and  demon- 
strations themselves.  To  correct  the  errors  which 
are  no^v  found  in  it,  and  bring  it  nearer  to  the  accu- 
racy with  which  it  ^^  as,  no  doubt,  at  first  written  by 
Euclid,  is  the  design  of  this  edition,  that  so  it  may 
be  rendered  more  useful  to  geometers,  at  least  to  be- 
ginners ^^  ho  desire  to  learn  the  investigatory  method 
of  the  ancients.  And  for  their  sakes,  the  composi- 
tions of  most  of  the  Data  are  subjoined  to  their  de- 
monstrations, that  the  compositions  of  problems  solv- 
ed by  help  of  the  Data  may  be  the  more  easily  made. 

Marinus  the  philosopher's  preface,  which,  in  the 
Greek  edition,  is  prefixed  to  the  Data,  is  here  left 
out,  as  being  of  no  use  to  understand  them.  At  the 
end  of  it  he  says,  that  Euclid  has  not  used  the  syn- 
thetical, but  the  analytical  method  in  delivering  them; 
in  which  he  is  quite  mistaken;  for,  in  the  analysis  of 
a  theorem,  the  thing  to  be  demonstrated  is  assumed 
in  the  analysis ;  but  in  the  demonstrations  of  the  Data, 
the  thing  to  be  demonstrated,  which  is,  that  some- 
thing or  other  is  given,  is  never  once  assumed  in  the 
demonstration,  from  which  it  is  manifest,  that  every 
one  of  them  is  demonstrated  synthetically;  though, 
indeed  if  a  proposition  of  the  Data  be  turned  into  a 
problem,  for  example  the  84th  or  85th  in  the  former 
editions,  which  here  are  the  85th  and  86th,  the  de- 
monstration of  the  proposition  becomes  the  anal}'sis 
of  the  problem. 

W^herein  this  edition  differs  from  the  Greek,  and 
the  reasons  of  the  alterations  from  it,  ^vill  be  shown 
in  the  notes  at  tlie  end  of  the  Data. 


(iBttcIiD^S  2Data, 


DEFINITIONS. 

I. 
SPACES,  lines,  and  angles,  are  said  to  be  given  in  magnitude, 
when  equals  to  them  can  be  found. 

II. 
A  ratio  is  said  to  be  given,  when  a  ratio  of  a  given  magnitude 
to  a  given  magnitude  which  is  the  same  ratio  with  it  can  be 
found. 

III. 
Rectilineal  figures  are  said  to  be  given  in  species,  which  have 
each  of  their  angles  given,  and  the  ratios  of  their  sides  given. 
IV. 
Points,  lines,  and  spaces,  are  said  to  be  given  in  position,  which 
have  always  the  same  situation,  and  which  are  either  actually 
exhibited,  or  can  be  found. 

A. 
An  angle  is  said  to  be  given  in  position  Avhich  is  contained  by 
straight  lines  given  in  position. 

V. 

A  circle  is  said  to  be  given  in  magnitude,  when  a  straight  line 

from  its  centre  to  the  circumference  is  given  in  magnitude. 

VI. 
A  circle  is  said  to  be  given  in  position  and  magnitude,  the  cen- 
tre of  which  is  given  in  position,  and  a  straight  line  from  it 
to  the  circumference  is  given  in  magnitude. 
VII. 
Segments  of  circles  are  said  to  be  given  in  magnitude,  when 
the  angles  in  them,  and  their  bases,  are  given  in  magnitude. 
VIII. 
Segments  of  circles  are  said  to  be  given  in  position  and  magni- 
tude,  when  the  angles  in  them  are  given  in  magnitude,  and 
their  bases  are  given  both  in  position  and  magnitude. 

IX. 
A  magnitude  is  said  to  be  greater  than  another  by  a  given  mag- 
nitude, when  this  given  magnitude  being  taken  from  it,  the 
Remainder  is  eq\ial  to  the  other  magnitude. 


;68 


EUCLID'S 


2. 


See  N. 


X. 

A  magnitude  is  said  to  be  less  than  afiother  by  a  given  magni- 
tude, when  this  given  magnitude  being  added  to  it,  the  whole 
is  equal  to  the  other  magnitude. 

1.  •  PROPOSITION  I. 

SeeN.  THE  ratios  of  given  magnitudes  to  one  another  is 

G'iven. 

Let  A,  B  be  two  given  magnitudes,  the  ratio  of  A  to  B  is 
given. 

Because  A  is  a  given  magnitude,  there  may 
a  1  def  *  ^^  found  one  equal  to  it ;  let  this  be  C  :  and 
dat.  because   B   is  given,  one  equal  to  it  may  be 

found  ;  let  it  be  D ;  and  since  A  is  equal  to  C, 
b  7.  5.  and  B  to  D ;  therefore  b  A  is  to  B,  as  C  to 
D ;  and  consequently  the  ratio  of  A  to  B  is 
given,  because  the  ratio  of  the  given  magni- 
tudes C,  D  which  is  the  same  with  it,  has  been  A  B  C  D 
found. 

PROP.  IL 

IF  a  elvcn  maQ:nitucIe  has  a  ^Ivtn  ratio  to  ano'ther 
magnitude,  "  and  if  unto  the  two  magnitudes  l^y 
"  which  the  c-iven  ratio  is  exhibited,  and  the  civen 
"  magnitude,  a  fou.rth  proportional  can  be  found  ;" 
the  otiicr  maicnitude  is  G:i'»en. 

Let  the  given  magnitude  A  have  a  given  ratio  to  the  mag- 
nitude B ;  if  a  fourth  proportional  can  be  found  to  the  three 
magnitudes  above  named,  B  is  given  in  magnitude 
Because  A  is  given,  a  magnitude  may  be 

a  1.  def.      found    equal  to  it  a;  let  this  be  C:  and  be- 
cause the  ratio  of  A  to  B  is  given,  a  ratio 
which  is  the  same  •v^'ith  it  may  be  found,  let 
this  be  the  ratio  of  the  given  magnitude  E       A    B    C    D 
to  the  given  magnitude  F  :  imto  tiie  magni-  E    F 

tudes  E,  F,  C  find  a  fourth  proportional 
D,  which,  by  the  hypothesis,  can  be  done. 
Wherefore, because  A  is  to  B,  as  E  to  F  ;  and 

b  11.  5.      as  E  to  F,  so  is  C  to  D  ;  A  is  b  to  B,  as  C  to 

*  The  figures  in  the  marg'in  sbow  the  number  of  the  propositions  iu 
the  other  euitloiis. 


DATA.  369 

D.  But  A  is  equal  to  C  ;  therefore  c  B  is  equal  to  D.  The  c  14.  5. 
magnitude  B  is  therefore  given  a  because  a  magnitude  D  equal  a  1.  def, 
to  it  has  been  found. 

The  limitation  within  the  inverted  commas  is  not  in  the 
Greek  text,  but  is  now  necessarily  added ;  and  the  same  must 
be  understood  in  all  the  propositions  of  the  book  which  depend 
Upon  this  second  proposition,  where  it  is  not  expressly  mention- 
ed.    See  the  note  upon  it. 


PROP.  III.  3 

IF  any  given  magnitudes  be  added  together,  their 
sum  shall  be  given. 

Let  any  given  magnitudes  AB,  BC  be  added  together,  their 
sum  AC  is  given. 

Because  AB  is  given,  a  magnitude  equal  to  it  may  be  founda;  a  1.  def. 
let  this  be  DE  :   and  because  BC  is  gi-  A  B  C 

ven,  one  equal  to  it  may  be  found  ;  let  ■ j — 

this  be  EF  :  wherefore,  because  AB  is 

equal  to  DE,  and  BC  equal  to  EF  ;  the  D  E         F 

whole  AC  is  equal  to  the  whole  DF  :  ij — 

AC    is   therefore   given,  because    DF 
has  been  found  which  is  equal  to  it. 


PROP.  IV.  4> 

IF  a  eiven  magnitude  be  taken  from  a  eiven  mai>:. 
nitude ;  the  remainina;  masmitude  shall  be  eiven. 

From  the  given  magnitude  AB,  let  the  given  magnitude  AC 
be  taken  ;  the  remaining  magnitude  CB  is  given. 

Because  Aii  is  given,  a  magnitude  equal  to  it  may  a  be  a  1.  def. 
found  ;  let  this  be  DE  :  and  because    .  PR 

AC  is  given,  one  equal  to  it  may  be  , 

found  ;  let  tiiis  be  DF  :  wherefore  be-  ' 

cause  AB  is  equal  to  DE,  and  AC  to  y~.  F         F 

Di' ;  the  remainder  C  'S  is  equal  to  the 
remainder  FE.       CB    is   therefore  '         ' 

given  a,  because  FE  which  is  equal  to  it  has  been  found. 

3  A 


3ro  ,  EUCLID'S 

12.  PROP.  V. 

See  N.  lY  qJ-  three  magnitudes,  the  first  together  with  the 
second  be  given,  and  also  the  second  together  -s^ith 
the  third;  either  the  first  is  equal  to  the  third,  or 
one  of  tliem  is  gTeater  than  the  other  by  a  given  mag- 
nitude. 

Let  AB,  BC,  CD  be  three  magnitudes,  of  which  AT5  together 
with  LC,  that  is  AC,  is  given  ;  and  also  BC  together  with  CD, 
that  is  ED,  is  given.  Either  A)'  is  equal  to  CD,  or  one  of  them 
is  greater  than  the  other  by  a  given  magnitude. 

I'ec-iuse  AC,  BD  are  each  of  them  given,  they  are  either 
equcil  to  one  another,  or  not  equal. 
First,  let  them  be  equal,  and  because       A      B  CD 

AC  is  equal  to  BD,  take  away  the  com-       [ ] 

mon  part  B  C;  therefore  the  remuin- 
der  A"  is  equal  to  the  remainder  CD. 

Rut  if  they  be  unequal,  let  AC  be  greater  than  BD,  and  make 
CE  equal  to  BD.     Therefore  CE  is  given,  because  BD  is  given. 
And  the  whole   AC  is  given  ; 
a  4.  dut.     therefore  a  AE  the  remainder  is  A      E  B  C         D 

given.  And  because  EC  is  equal  1 1     ■        •! 

io  BD,  by  taking  liC  from  both, 

the  remainder  EP  is  equal  to  the  remainder  CD.  And  AE  is 
given  ;  wherefore  AB  exceeds  EB,  that  is  CD  by  the  given 
magnitude  AE. 

5.  PROP.  VL 

See  N.  IF  a  magnitude  has  a  given  ratio  to  a  part  of  it,  it 

shall  also  have  a  given  ratio  to  the  remaining  part 
of  it. 

Let  the  magnitude  AB  have  a  given  ratio  to  AC  a  part  of  it ' 

it  has  also  a  given  ratio  to  the  remainder  BC. 

Because  the  ratio  of  Ali  to  AC  is  given,  a  ratio  maybe 
a  2.  dcf      found  a  which  is  the  same  to  it :  let  this  be  the  ratio  of  DE 

a  given  magnitude  to  the  given  magni-    .  PR 

tude  DF.  And  because  DE,  DF  are  gi-  ^ , 

b  4.  dat.     ven,  the  remainder  FE  is  b  given  :  and 

because  AB  is  to  AC,  as  DE  to  DF,  by   .^  -p         p 

c  E.  5.        conversion c  AB  is  to  BC,  as  Di-.  to  EF. ^ 

Therefore  the  raiio  of   AB  to  BC  is 

given,  because  theratioof  the  given  magnitudes  DE,  EF,  which 

is  the  same  with  it,  has  been  found. 


DATA.  wi 


Cor,  Froin  this  it  follows,  that  the  parts  AC,  CB  have  a  given 
ratio  to  one  another:  because  as  AB  to  BC,  so  is  DE  to  EF  ; 
by  division d,  AC  is  to  CB,  as  DF  to  FE  :  and  DF,  FE  ared  17.  5. 
given  ;  therefore  a  the  ratio  of  AC  to  CB  is  given.  a  2.  def. 


PROP.  VII.  6. 

IF  two  magnitudes  \\'hich  have  a  given  ratio  to  one  See  n. 
another,  be  added  together;  the  vrhoie  magnitude  skill 
have  to  each  of  them  a  gi\"en  ratio. 

Let  the  magnitudes  AB,  BC  which  have  a  given  ratio  to  one 
another,  be  added  together  ;  the  whole  AC  has  to  each  of  the 
magnitudes  AB,  BC  a  given  ratio. 

Because  the  ratio  of  AB  to  BC  is  given,  a  ratio  may  be 
found  a  wnich  is  the  same  with  it;  let  this  be  the  ratio  of  the  a  2.  def. 
given  magnitudes  DE,  EF  :    and  be- 
cause  DE,   EF  are  given,  the  whole    A  B  C 

DF  is  given b  :  and  because  as  AB  to    [ b  3.  dat. 

BC,  so  is  DE  to  EF ;  by  composition    D  E  F 

cAC  is  to  CB,  as  DF  to  FE;  and  by 1 «  18.  5. 

conversiond,  AC  is  to  AB,  as  DF  to  d  E.  5. 

DE :  wherefore  because  AC  is  to  each  of  the  magnitudes  AB, 
BC,  as  DF  to  each  of  the  others  DE,  EF  ;  the  ratio  of  AC  tp 
each  of  the  magnitudes  AB,  BC  is  givena. 

PROP.  VIII.  r. 

IF  a  given  magnitude  be  divided  into  two  paits  See  Note. 
which  have  a  given  ratio  to  one  another,  and  if  a 
fourth  proportional  can  be  foimd  to  the  sum  of  the 
two  magnitudes  by  which  the  given  ratio  is  exhibited, 
one  of  them,  and  the  given  magnitude ;  each  of  the 
parts  is  given. 

Let  the  given  magnitude  AB  be  divided  into  the  parts  AC, 
CB  which  have  a  given  ratio  to  one  another ;  if  a  fourth  pro- 
portional can  be  found  to  the  above 

named  magnitudes  ;  AC  and  CB  are  A  C  B 
each  of  them  given.  t 

Because  the  ratio  of  AC  to  CB  is  D  F  E 

given  ;  the    ratio  of   AB    to    BC   is — ; 

given  a;    therefore  a  ratio  which  is  a  7.  dat. 


372 


EUCLID'S 


b  2.  clef,    tlie   same   Avith  it   can  be   foundb,   let  this  be   the   ratio  of 
the  given  magnitudes  DE,   EF  :  and 
because  the  given  magnitude  AB  has  A  C  B 

to  BC  the  given  ratio  of  DE  to  EF,  if  ]- 

unto   DE,  EF,   AB  a  fourth  propor-  D                  I'     E 
tional  can  be  found,  this  which  is  BC    1 

c  2.  dat.    is  given  c  ;  and  because  AB  is  given 

d  4.  dat.    the  other  part  AC  is  given  d. 

In  the  same  manner,  and  with  the  like  limitation,  if  the  dif- 
ference AC  of  two  magnitudes  AB,  BC  which  have  a  given  ra- 
tio, be  given  ;  each  of  the  magnitudes  AB,  BC  is  given. 


8. 


PROP.  IX. 


MAGNITUDES  whch  h^\e  pjiven  ratios  to  the 
same  magnitude,  have  also  a  given  ratio  to  one  ano- 
ther. 


Let  A,  C  have  each  of  them  a  given  ratio  to  B  ;  A  has  a 
given  ratio  to  C. 

Because  the  ratio  of  A  to  B  is  given,  a  ratio  which  is  the 
a  2,  def.  same  to  it  may  be  found  a ;  let  this  be  the  ratio  of  the  given 
magnitudes  D,  E  :  and  because  the  ratio  of  B  to  C  is  given,  a 
ratio  which  is  the  same  Avith  it  may  be  founds  ;  let  this  be  the 
ratio  of  the  given  magnitudes  F,  G : 
to  F,  G,  E  find  a  fourth  propor- 
tional H,  if  it  can  be  done;  and 
because  as  A  is  to  B,  so  is  D  to  E  ; 
and  as  V>  to  C,  so  is  (F  to  G,  and 
so  is)  E  to  H  ;  ex   xcjuali^  as  A  to 

C,  so  is  D  to  H  ;  therefore  the  ra-   A      li      C      D      E 
tio  of  A  to  C  is  given  a,  because  the  ^ 

ratio  of  the  given  magnitudes  D  and 
H,  which  is  the  same  with  it,  has 
been  found  :  but  if  a  fourth  propor- 
tional to  F,  G,  E  cannot  be  found, 
then  it  can  only  be  said  that  the  ratio  of  A  to  C  is  compounded 
of  the  ratios  of  A  to  H,  and  B  to  C,  that  is,  of  the  given  ratios 
of  D  to  i'^,  and  F  to  Ci . 


H 
G 


DATA.  373 


PROP.  X. 


IF  two  or  more  magnitudes  have  gi^'en  ratios  to  orx 
another,  and  if  they  have  given  ratios,  thougli  they  be 
not  the  same,  to  some  other  magnitudes  ;  these  other 
magnitudes  sh.Jl  also  have  given  ratios  to  one  another. 

Let  two  or  more  magnitudes  A,  B,  C  have  given  ratios  to 
one  another  ;  and  let  them  have  given  ratios,  though  they  be 
not  the  same,  to  some  other  magnitudes  D,  E,  F  :  the  magni- 
tudes D,  E,  F  have  given  ratios  to  one  another. 

Because  the  ratio  of  A  to  B  is  given,  and  likewise  the  ratio 
of  A  to  D  ;  therefore  the  ra- 
tio of  D  to  B  is  givena;  but     A D 

the  ratio  of  B  to  E  is  given,     B E  — 

therefore  a  the  ratio  of  D  to      C  — —  F a  9.  dat. 

E  is  given  :   and  because  the 

ratio  of  B  to  C  is  given  :  and  also  the  ratio  of  B  to  E  ;  the  ratio 
of  E  to  C  is  givena ;  and  the  ratio  of  C  to  F  is  given ;  where- 
fore the  ratio  of  E  to  F  is  given  :  D,  E,  F  have  therefore  given 
ratios  to  one  another. 


PROP.  XI.  22. 

IF  two  magnitudes  have  each  of  them  a  given  ratio 
to  another  magnitude,  both  of  them  together  shall  ha\'e 
a  giA^en  ratio  to  that  other. 

Let  the  magnitudes  AB,  BC  have  a  given  ratio  to  the  mag- 
nitude D  ;   AC  has  a  given  ratio  to  the  same  D. 

Because  All,  BC  have  each  of 
them  a  given  ratio  to  D,  the  ratio  B 

of  AB  to  BC  is  givena  ;  and  by  A 1 C  a  9.  dat. 

composition,  the  ratio  of  AC  to  D 

CB  is  given b  :  but  the  ratio  of  b  7.  dat. 

BC  to  D  is  given  ;  therefore  a  the  ratio  of  AC  to  D  is  given. 


S74  EUCLID'S 


PROP.  XII. 


See  N.  IF  the  whole  ha^-e  to  the  whole  a  gi\en  ratio,  and 

the  parts  have  to  the  parts  given,  but  not  the  same, 
ratios,  ever}'  one  of  them,  ■whole  or  part,  shall  liave  to 
e^-ery  one  a  given  ratio. 

Let  the  whole  AB  have  a  given  ratio  to  the  whole  CD,  and 
the  parts  AE,  EB  have  given,  but  not  the  same,  ratios  to  the 
parts  CF,  FD,  every  one  shall  have  to  every  one,  whole  or  part, 
a  given  ratio. 

Because  the  ratio  of  AE  to  CF  is  given,  as  AE  to  CF,  so 
make  AB  to  CG  ;  the  ratio  therefore  of  AB  to  CG  is  given  ; 
wherefore  the   ratio  of  the  remainder  EB  to  the  remainder 

:i  19.  5.     pQ  j^g  given,  because  it  is  the  same  a  with  the  ratio  of  AB  to 
CG  ;  and  the  ratio  of  EB  to  FD  is 
given,  wherefore  the  ratio  of  FD 

b  9   dat    to  FG  is  given  b  ;  and  by  conver- 
sion,  the  ratio  of   FD  to  DG  is 

c  6.  dat.  given c  ;  and  because  AB  has  to 
each  of  the  magnitudes  CD,  CG  a 
given  ratio,  the  ratio  of  GD  to  CG  is  given b  ;  and  therefore  c 
the  ratio  of  CD  to  DG  is  given  :  but  the  ratio  of  GD  to  DF  is 
given,  whereforeb  the  ratio  of  CD  to  DF  is  given,  and  conse- 

d  cor.  6.   quently  (1  the  ratio  of  CF  to  FD  is  given  ;  but  the  ratio  of  CF  to 

e'Vo   dat  ^^  given,  as  also  the  ratio  of  FD  to  EB,  whereforee  the  ra- 

tio of  AE  to  EB  is  given;  as  also  the  ratio  of  AB  to  each  of 

f  7.  dat.    thenif :  the  ratio  therefore  of  every  one  to  every  one  is  given. 

24  PROP.  XIII. 

See  N.  jY  ^^^^^  |-j,j,_^  Q.|>  three  proportional  straight  lines  has  a 

\'en  ratio  to  the  t 
ratio  to  the  !::eeor.d 


A 

E 

B 

C 

F 

G 

D 

gi\'en  ratio  to  the  third,  the  lir.st  sliali  also  have  a  f^iven 


Let  A,  B,  C  be  three  proportional  straight  Unes,  that  is,  as  A 
to  B,  so  is  B  to  C  ;  if  A  has  to  C  a  given  ratio,  A  shall  also  have 
to  B  a  given  ratio. 

Because  the  ratio  of  A  to  C  is  given,  a  ratio  which  is  the 
a  2.  dcf.  5,ame  with  it  may  be  foirnda  ;  let  this  be  the  ratio  of  the  given 
fa  13.  6.     </a-aighc   lines   D,   E;    and  !)ctween    Band   E  find  a  bineai> 


DATA. 


375 


proportional  F;   therefore  the  rectangle  contained  by  D   and 
E  is  equal  to  the  square  of  F,  and  the   rect- 
angle D,  E  is  given,  because  its  sides  D,  E  are 
given  ;  wherefore  the   square  of  F,  and   the 
straight  line  F  is  given  :  and  because  as  A  is 
to  C,  so  is  D  to  E  ;  but  as  A  to  C,  so  is  c  the 
square  of  A  to  the  square  of  B  ;  and  as  D  to 
E,  so  is  c  the  square  of  D  to  the  square  of  F  :     A      B      C 
therefore  the  square  d  of  A  is  to  the  square  of 
B,   as  the   square  of  D  to  the  square  of  F  :     D      F      E 
as  therefore  e  the  straight  line  A  to  the  straight 
line  B,  so  is  the  straight  line  D  to  the  straight 
line  F  :  therefore  the  ratio  of  A  to  B  is  given  a, 
because  the  ratio  of  the  given  straight  lines  D, 
F  which  is  the  same  with  it  has  been  found. 


c  2  cor. 
20.  6. 

d  11.  5. 

e  22.  6. 
a  2.  def 


PROP.  XIV. 


A. 


IF  a  magnitude  together  with  a  given  magnitude  see  n. 
has  a  given  ratio  to  another  magnitude  ;  the  excess 
of  tiiis  other  magnitude  above  a  given  magnitude  has 
a  s:i\'en  ratio  to  the  first  magnitude  :  and  if  the  ex- 
cess  of  a  magnitude  above  a  given  magnitude  has  a 
given  ratio  to  another  magnitude :  this  other  magni- 
tude together  with  a  given  magnitude  has  a  .given 
ratio  to  the  first  magnitude. 


Let  the  magnitude  AB  together  with  the  given  magnitude 
BE,  that  is  AE,  have  a  given  ratio  to  the  magnitude  CD ;  the 
excess  of  CD  above  a  given  magnitude  has  a  given  ratio  to  AB. 

Because  the  ratio  of  AE  to  CD  is  given,  as  AE  to  CD,  so 
make  BE  to  FD  ;  therefore  the  rutio  of  BE  to  FD  is  given,  and 
BE  is  given;  wherefore  FD  is  gi- 
ven a  :  and  because  as  AEto  CD,  so 
is  BE  to  FD,  the  remainder  AB  is  b 
to  the  remainder  CF,  as  AE  to  CD  : 
but  the  ratio  of  AE  to  CD  is  given, 
therefore  the  ratio  of  AB  to  CF  is 
given  ;  that  is,  CFthe  excess  of  CD  above  the  given  magnitude 
FD  has  a  given  ratio  to  AB. 

Next,  Let  the  excess  of  the  magnitude  AB  above  the  given 
magnitude  BE,  that  is,  let  AE  have  a  given  ratio  to  the  mag-- 


A 


E 


a  2.  dat 
b  19.  5. 


D 


376  EUCLID'S 

nitude  CD  :  CD  together  with  a  given  magnitude  has  a  given 
ratio  to  AB. 

Becuuse  the  ratio  of  AE  to  CD  is   given,  as   AE  to  CD,  so 
make  BE  to  FD  ;  therefore  the  ratio  of 
BE  to   FD  is  given,  and  BE  is  given,     A  E         B 

a2.dat.     wherefore  FD  is  given  a.     And  because 1 

as  AE  to  CD,  so  is  BE  to  FD,  AB  is 

c  12.  5.      to  CF,  as  c  AE  to  CD  :  but  the  ratio  of    C  D     F 

AE  to  CD  is  given,  therefore  the  ratio     il 

of  AB  to  CF  is  given  :  that  is,  CF  which  is  equal  to  CD,  toge- 
ther with  the  given  magnitude  DF,  has  a  given  ratio  to  AB. 

B  PROP.  XV. 

See  Note.  IF  a  magnitude,  together  with  that  to  which  ano- 
ther mag:iitude  has  a  given  ratio,  be  given ;  the  i:.um 
of  tliirs  other,  and  that  to  wliich  the  hrst  magnitude 
has  a  given  ratio,  is  given. 

Let  AB,  CD  be  two  magnitudes,  of  Avhich  AB  together  with 

BE,  to  which  CD  has  a  given  ratio,  is  given ;  CD  is  given, 

together  with  that  magnitude  to  which  AL'-  has  a  given  ratio. 
Because  the  ratio  of  CD  to  iiE  is   given,  as  BE  to  CD,  so 

make  AE  to  FD  ;  therefore  the  ratio  of  AE  to  FD  is  given,  and 
a  2  dat.       AE  is  given,  wherefore  a  FD  is  given  :     .  R  F 

and  because  as  BE  to  CD,   so  is  AE  to  , 

b  Cor.  19.   FD:  ABisb  to  FC,  as  BE  to  CD:  and 

5-  the  ratio  of  BE  to  CD  is  given,  where-  F         C       D 

fore  the  ratio  of  AB  to  FC  is  given  :  and  !— 

FD  is  given,  that  is,  CD  together  v/ith 

FC,  to  which  AB  has  a  given  ratio,  is  given. 


10. 


PROP.   XVL 


See  Note.  IF  the  excess  of  a  magnitude  above  a  given  magni- 
tude, has  a  given  ratio  to  another  magnitude  ;  the 
excess  of  both  together  above  a  given  magnitude  shall 
have  to  that  other  a  given  ratio  :  and  if  the  excess  of 
two  magnitudes  together  above  a  giA'en  magnitude, 
has  to  one  of  them  a  given  ratio ;  either  the  excess 
of  the  other  above  a  gi^  en  magnitude  has  to  that  one 
a  given  ration,  or  the  other  is  gi\'en  together  with  the 
magnitude  to  vdiich  that  one  has  a  given  ratio. 


DATA.  3?r 

Let  the  excess  of  the  magnitude  AB  above  a  given  magni- 
tude, have  a  given  ratio  to  the  magnitude  BC  ;  the  excess  of 
AC,  both  of  them  together,  above  the  given  magnitude,  has  a 
given  ratio  to  BC. 

Let  AD  be  the  given  magnitude,  the  excess  of  AB  above 
which,  viz.  DB  has  a  given  ratio    .  D        R  P 

to  BC :  and  because  DB  has  a  giv-  ^ 

en  ratio  to  BC,  the  ratio  of  DC  to  ' 

CB  is  given  a,  and  AD  is  given  ;  therefore  DC,  the  excess  of  a  7.  dat. 

AC  above  the  given  magnitude  AD,  has  a  given  ratio  to  BC. 

Next,  let  the  excess  of  tAVo  magnitudes  AB,  BC  together, 
above  a  given  magnitude,  have  to       .  D        R        F      r 

one  of  them  BC  a  given  ratio ;  ei- ^^ i___ 

ther  the  excess  of  the   other  of  1  i         I 

them  AC  above  the  given  magnitude  shall  have  to  BC  a  given 
ratio ;  or  AB  is  given,  together  with  the  magnitude  to  which 
BC  has  a  given  ratio. 

Let  AD  be  the  given  magnitude,  and  first  let  it  be  less  than 
AB ;  and  because  DC  the  excess  AC  above  AD  has  a  given  ra- 
tio to  BC,  DB  hash  a  given  ratio  to  BC;  that  is,  DB  the  excess b  Cor.  6. 
of  AB  above  the  given  magnitude  AD,  has  a  given  ratio  to  BC.^** 

But  let  the  given  magnitude  be  greater  than  AB,  and  make 
A\r.  equal  to  it ;  and  because  EC,  the  excess  of  AC  above  A  P., 
has  to  BC  a  given  ratio,  BC  hasc  a  given  ratio  to  B'vl ;  and  be-c  6.  dat. 
cause  AK  is  given,  AB  together  with  BE,  to  which  BC  has  a 
given  ratio,  is  given. 


PROP.  XVIL  11. 

IF  the  excess  of  a  magnitude  above  a  given  magni-  See  Note. 
tude  has  a  given  ratio  to  another  magnitude  ;  the  ex- 
cess of  the  same  first  magnitude  above  a  given  mag- 
nitude, shall  have  a  given  ratio  to  both  the  magnitudes 
together.  And  if  the  excess  of  either  of  two  magni- 
tudes above  a  gi\'en  magnitude  lias  a  gi\'en  ratio  to 
JDOth  magnitudes  together:  the  excess  of  the  same 
aboA-e  a  given  magnitude  shall  have  a  gi\'en  ratio  to 
the  other. 

Let  the  excess  of  the  magnitude  AB  above  a  given  magni- 
tude have  a  given  ratio  to  the  magnitude  BC  ;  the  excess  of  AB 
above  a  given  magnitude  has  a  given  ratio  to  AC. 

3  B 


378  EUCLID'S 

Let  AD  be  the  given  magnitude  ;  and  because  ll/'B,  the  ex- 
cess of  AB  above  AD,  has  a  given  ratio  to  EC  ;  the  ratio  of  DC 

a  7.  dat,  to  DB  is  given  a  ;  make  the  ratio  of  AD  to  DE  the  same  with 
this  ratio ;  therefore  the  ratio  of  .  F  D  B  C 
AiJ  to  DE  is  given:  and  AD  is  __^^___r 

-b  2.  dat.    given,  wherefore b  D',,  and   the  III 

remainder  AE  are  given:  and  because  as  DC  to  DB,  so  is  AD 

c  12.  5.  to  DE,  AC  isc  to  EB,  as  DC  to  DB;  and  the  ratio  of  DC  to 
DB  is  given;  wherefore  the  ratio  of  AC  to  LB  is  given:  and 
because  the  ratio  of  EB  to  AC  is  given,  and  that  AE  is  given, 
therefore  EB  the  excess  of  AB  above  the  given  magnitude  AE, 
has  a  given  ratio  to  AC. 

Next,  Let  the  excess  of  AB  above  a  given  magnitude  have  a 
given  ratio  to  AB  and  BC  together,  that  is,  to  AC  ;  the  excess 
of  A3  above  a  given  magnitude  has  a  given  ratio  to  BC. 

Let  AE  be  the  given  magnitude;  and  because  EB  the  excess 
of  AB  above  AE  has  to  AC  a  given  ratio,  as  AC  to  EB,  so  make 

<l  6.  dat.  AD  to  DE;  therefore  the  ratio  of  AD  to  DL  is  given,  as  alsod 
the  ratio  of  AD  to  AE  :  and  AE  is  given,  wherefore  b  AD  is 
given  :  and  because,  as  the  whole  AC,  to  the  whole  EB,  so  is 

e  19.  5.  AD  to  DE,  the  remainder  DC  ise  to  the  remainder  DB,  as  AC 
to  EB  ;  and  the  ratio  of  AC  to  EB  is  given ;  wherefore  the  ratio 

f  Cor.  6.    of  DC  to  DB  is  given,  as  alsof  the  ratio  of  DB  to  BC  :  and  AD 

dat.  is  given  ;  therefore  DB,  the  excess  of  AB  above  a  given  mag- 

nitude AD,  has  a  given  ratio  to  BC. 


14.  PROP.  XVIIL 

IF  to  each  of  two  mao-nitudes,  wlilcli  ha^e  a  .f>lven 
ratio  to  one  another,  a  given  magnitude  be  added;  the 
^vhole  shall  either  have  a  given  ratio  to  one  another,  or 
tb.e  excess  of  one  of  them  abo^e  a  .G:iven  niae^nitude 
shall  ha\^e  a  given  ratio  to  the  other. 


b' 


Let  the  two  magnitudes  AB,  CD  liave  a  given  ratio  to  one 

another,  and  to  AB  let  the  given  magnitude  BE  be  added,  and 

the  given  magnitude  DF  to  CD  ;  the  wholes  AE,  CF  either 

have  a  given  ratio  to  one  another,  or^the  excess  of  one  of  them 

a  1.  flat,    above  a  given  magnitude  has  a  given*  ratio  to  the  othu"'^ 

Because  li£,  DF  are  each  of  them  given,  their  ratio  is  given. 


DATA.  379 


ind  if  this  ratio  be  the  same  with     A           B 
the  ratio  of  x\B  to  CD,  the  ratio  of      1- 


AE  to  CF,  which  is  the  same  b  with     CD  F  b  12.  5. 

the  given  ratio  of  AB  to  CD,  shall       1 

be  given. 

But  if  the  ratio  of  BE  to  DF  be  not  the  same  with  the  ratio 
of  AB  to  CD,  either  it  is  greater  than  the  ratio  of  AB  to 
CD,  or,  by  inversion,  the  ratio  of  DF  to  BE  is  greater  than 
the  ratio  of  CD   to  AB :  first,  let      .        ^  ^  ^ 

the  ratio  of  BE  to  DF  be  greater 

than  the  ratio  of  AB  to  CD  ;  and  as     "Z         ^^T  ^"7i 

AB  to  CD,  so  make  BG  to  DF  ;     ___7 

therefore  the  ratio  of  BG  to  DF  is 

given  ;  uud  DF  is  given,  thereforec  BG  is  given  :  and  because  c  2.  dat. 
BE  has  a  greater  ratio  to  DF  than  (AB  to  CD,  that  is,  than) 
BG  to  DF,  BE  is  greaterd  than  BG  ;  and  because  as  AB  to  CD,^  ^^-  ^• 
so  is  BG  to  DF ;  therefore  AG  isb  to  CF,  as  AB  to  CD  :  but 
the  ratio  of  AB  to  CD  is  given,  wherefore  the  ratio  of  AG  to 
CF  is  given ;  and  because  BE,  BG  are  each  of  them  given,  GE 
is  given :  therefore  AG  the  excess  of  AE  above  a  given  magni- 
tude GE,  has  a  given  ratio  to  CF.  The  other  case  is  demon- 
sti'ated  in  the  same  manner. 


PROP.  XIX.  15. 


IF  li^om  each  of  two  magnitudes,  which  ha\e  a 
Hven  ratio  to  one  another,  a  s-ivgw  raa^cnitiide  be 
taken,  the  remainders  shall  either  have  a  given  ratio 
to  oii.e  another,  or  the  excess  of  one  of  them  above  a 
givea  magnitude,  sliall  have  a  gi\en  ratio  to  the  other. 

Let  the  magnitudes  AB,  CD  have  a  given  ratio  to  one  ano- 
ther, and  from  AB  let  the  given  magnitude  AE  be  taken,  and 
from  CD,  the  given  magnitude  CF  :  the  remainders  EB,  FD 
shall  either  have  a  given  ratio  to  one  another,  or  the  excess 
of  one    of  them   above  a  given      a  p  Tt 

magnitude    shall    have    a  given  

ratio  to  the  other.  „       '       p. 

Because  AE,  CF  are  each  of  , 

them  given,  their  ratio  is  given  a:  a  1.  dat. 

and  if  this  ratio  be  the  same  with  the  ratio  of  AB  to  CD,  the 


5^  EUCLID'S 

ratio  of  the  remainder  EB  to  the  remainder  FD,  which  is  the 
b  19.  5.     sameb  with  the  given  ratio  of  AB  to  CD,  shall  be  given. 

But  if  the  ratio  of  AB  to  CD  be  not  the  same  with  the  ratio 
of  AE  to  CF,  either  it  is  greater  than  the  ratio  of  AE  to  CF, 
or,  by  inversion,  the  ratio  of  CD  to  AB  is  greater  than  the  ra- 
tio of  CF  to  AE.  First,  let  the  ratio  of  AB  to  CD  be  greater 
than  the  ratio  of  AE  to  CF,  and  as  AB  to  CD,  so  make  AG  to 
CF ;  therefore  the  ratio  of  AG  to    .  v  c  H 

CF  is  given,  and  Ct   is   given,     

c  2.  dat.    wherefore  c  AG    is    given  :    and  p  '  p    '~T| 

because  the  ratio  of  AB  to  CD,     

that  is,  the  ratio  of  AG  to  CF,  ' 

d  10.  5.  is  greater  than  the  ratio  of  AE  to  CF  ;  AG  is  greater  d  than 
AE  :  and  AG,  AE  are  given,  therefore  the  remainder  EG  is 
given;  and  as  AB  to  CD,  so  is  AG  to  CF,  and  so  isb  the  re- 
mainder GB  to  the  remainder  FD  ;  and  the  ratio  of  AB  to  CD 
is  given  :  wherefore  the  ratio  of  GB  to  FD  is  given  ;  therefore 
GB,  the  excess  of  EB  above  a  given  magnitude  EG,  has  a 
given  ratio  to  FD.  In  the  same  manner  the  other  case  is  de- 
monstrated. 


16.  PROP.  XX. 

IF  to  one  of  t^vo  magnitudes  which  have  a  given 
ratio  to  one  another,  a  given  magnitude  be  added, 
and  from  the  other  a  given  magnitude  be  taken ;  the 
excess  of  tb.e  sum  above  a  sriven  magnitude  shall  have 

o  o 

a  given  ratio  to  the  remainder. 

Let  the  two  magnitudes  AB,  CD  have  a  given  ratio  to  one 
another,  and  to  AB  let  the  given  magnitude  EA  be  added,  and 
from  CD  let  the  given  magnitude  CF  be  taken  ;  the  excess  'of 
the  sum  EB  above  a  given  magnitude  has  a  given  ratio  the  re- 
mainder FD. 

Because  the  ratio  of  AB  to  CD  is  given,  make  as  AB  to 

CD,  so  AG  to  CF  :  therefore  the  ratio  of  AG  to  CF  is  given, 

a  2.  d  t.    and  CF  is  given,  wherefore  a  AG  p  a  C  P 

is  given ;  and  E A  is  given,  there-  . ^ 

fore  the  whole  EG  is  given :  and  Z  '  y  t> 

because  as  AB  to  CD,  so  is  AG t         , 

h  19.  5.     tQ  (iy_^  ^yy^  gQ  jgb  the  remainder 

GB  to  the  i-emainder  FD  ;  the  ratio  of  GB  to  FD  is  given. 
And  EG  is  given,  therefore  GB,  the  excess  of  the  sum  EB 


DATA'.  f        381 

above  the  given  magnitude  EG,  has  a  given  ratio  to  the  remain- 
der FD. 


PROP.  XXL  C. 

IF  two  magnitudes  have  a  given  ratio  to  one  ano-  See  n. 
ther,  if  a  given  magnitude  be  added  to  one  of  them, 
and  the  other  be  taken  from  a  gi\en  magnitude ;  tlie 
sum,  together  with  the  magnitude  to  ^\  hich  the  re- 
mainder has  a  given  ratio,  is  given  :  and  the  remain- 
der is  given  together  with  die  magnitude  to  which 
the  sum  has  a  eiven  ratio. 

o 

Let  the  two  magnitudes  AB,  CD  have  a  given  ratio  to  one 
another  ;  and  to  AB  let  the  given  magnitude  BE  be  added,  and 
let  CD  be  taken  from  the  given  magnitude  FD:  the  sum  AF. 
is  given,  together  with  the  magnitude  to  which  the  remainder 
FC  has  a  given  ratio. 

Because  the  ratio  of  AB  to  CD  is  given,  make  as  AB  to  CD, 

so  i^B  to  FD  :  therefore  the  ratio  of  GB  to  FD  is  given,  and 

Fi)  is  ariven,  wherefore  GB  is     ^^      *  n         t- 

J  T.1-    •       •          4-1        ^     A                BE  a  2.  dat, 

given  a;    and  BE   is  given,  the     ^__ , 

whole  G  E  is  therefore  given :  and 

because  as  AB  to  CD,  so  is  GB     v  r  D 

toFD,andsoisbGAtoFC;the     ____J J       b   19.  5, 

ratio  of  GA  to  FC  is  given  :  and 

AEl  together  with  G  A  is  given,  because  GE  is  given  ;  therefore 

the  sum  AE  together  with  GA,  to  which  the  remainder  FC  has 

a  given  ratio,  is  given.  The  second  part  is  manifest  from  prop.  15. 


PROP.  XXII.  D. 

IF  two  magnitudes  ha^v^e  a  eiven  ratio  to  one  anc-  See  n. 
ther,  if  fiom  one  of  them  a  given  magnitude  be  taken, 
and  the  other  be  taken  from  a  given  mxagniuide  ;  each 
of  the  remainders  is  given,  together  with  the  magni- 
tude to  "v^  hich  the  other  remainder  has  a  given  ratio. 

Let  the  two  magnitudes  AB,  CD  have  a  given  ratio  to  one 
^inother,  and  from  AB  let  the  given  magnitude  AE  be  taken, 


'3S2  EUCLID'S 

and  let  CD  be  taken  from  the  given  magnitude  CF  :  the  remain- 
der EB  is  given,  together  with  the  magnitude  to  which  the 
other  remainder  DF  has  a  given  ratio. 

Because  the  ratio  of  AB  to  CD  is  given,  make  as  AB  to  CD, 
so  AG  to  CF  :  the  ratio  of  AG  to  CF  is  therefore  given,  and 

a  2.  dut.     CF  is  given,  wherefore  a  AG  is    .  p-   p  ^. 

given ;    and  AE  is  given,   and  ' 

therefore  the  remainder  EG  is  '  . 

given  ;  and  because  as  AB  to  CD,  p  "D  F 

b  19.  5.       so  is  AG  to  CF  :  and  so  is  b  the 


remainder  BG  to  the  remainder  ^ 

DF  ;  the  ratio  of  BG  to  DF  is  given:  and  FB  together  with 
BG  is  given,  because  EG  is  given :  therefore  the  remainder  Y  B 
together  with  BG,  to  which  DF  the  other  remainder  has  a  given 
ratio,  is  given.  The  second  part  is  plain  from,  this  and  prop.  15. 


20.  PROP.  XXIII. 

See  N.  I^"  from  two  given    magnitudes  there  be  taken 

maG-nitudcs  which  have  a  eiven  ratio  to  one  another, 
the  remainders  shall  either  have  a  e;iven  ratio  to  cne 
another,  or  the  excess  of  one  of  them'  above  a  given 
magnitude  shall  have  a  given  ratio  to  the  other. 

Let  AB,  CD  be  two  given  magnitudes,  and  from  them  let  the 
magnitudes  AE,  CF,  wl.ich  have  a  given  ratio  to  one  another,  be 
taken  ;  the  remainders  EB,  FD  either  have  a  given  ratio  to  one 
another,  or  the  excess  of  one  of  them  above  a  given  magnitude 
has  a  given  rfitio  to  the  other. 

Because  AB,  CD  are  each  of  A  E  B 

them  given,  the  ratio  of  AB  to  ■ -1 

CD  is   given  :   and  if  this  ratio 

be  the  same  with  the  ratio  of  AE  C  F       D 

to  CF)  then  the    remahider  EB 1 

a  19.  5.      ^^^s  a  the  same  given  ratio  to  the 
remainder  FD. 

But  if  the  ratio  of  AB  to  CD  be  not  the  same  with  the  ra- 
tio of  AE  to  CF,  it.  is  either  greater  than  it,  or,  by  inversion, 
the  ratio  of  CD  to  AB  is  greater  than  the  ratio  of  CF  to  AE  : 
first,  let  the  ratio  of  AB  to  CD  be  greater  than  the  ratio  of 
AE  to  CF ;  and  as  AE  to  CF,  so  make  AG  to  CD  ;  there- 
fore the  ratio  of  AG  to  CD  is  given,  because  the  ratio  of 
b2.dat.     AE  to  CF  is  G-iven  ;    and  CD  is  civen.  wherefore I>  AG  i^ 


DATA.  3i 

given  ;  and  because  the  ratio  of  AB  to  CD  is  greater  than  the 
ratio  of  (AE  to  CF,  that  is,  than    .  V  C      ^ 

the  ratio  of)  AG  to  CD  ;   A  P.  is  ^_ , ^      ^ 

greater  c  than  AG  :  and  AP,AG  "  '  c  10. 5. 

are  given  ;  therefore  tlie  remain-  p  F       n 

der  ;"/G  is  given  :  and  because  as 

AE  to  CF,  so  is  AG  to  CD,  and  ' 

so  is  a  EG  to  DF  ;  the  ratio  of  EG  to  FD  is  given  :  and  GB  is  a  19-  5. 
given ;  therefore  EG,  the  excess  of  EB  above  a  given  magni- 
tude GB,  has  a  given  ratio  to  FD.     The  other  case  is  shown  in 
t^he  same  wav. 


PROP.  XXIV.  13. 

IF  there  be  three  magnitudes,  the  first  of  which  see  n. 
has  a  given  ratio  to  the  second,  and  the  excess  of  the 
second  above  a  gi^^en  magnitude  has  a  given  ratio  to 
the  third ;  the  excess  of  the  first  above  a  given  mag- 
nitude shall  also  have  a  given  ratio  to  the  third. 

Let  AB,  CD,  E,  be  the  three  magnitudes  of  v/hich  AB  has 
a  given  ratio  to  CD  ;  and  the  excess  of  CD  above  a  given  mag- 
nitude has  a  given  ratio  to  E :  the  excess  of  AB  above  a  given 
magnitude  has  a  given  ratio  to  E. 

Let  CF  be  the  given  magnitude,  the  excess  of  CD  above 
which,  viz.  FD  has  a  given  ratio  to  E :  and  because  the  ratio 
of  AB  to  CD  is  given,  as  AB  to  CD,  so  make  A 
AG  to  CF ;  therefore  the  ratio  of  AG  to  CF 
is  given ;  and  CF  is  given,  wherefore  a  AG  is   p 
given  :  and  because   as  AB  to  CD,  so  is  AG       "* 
to  CF,  and  so  is  b  QB  to  FD  ;  the  ratio  of  GB 
to  FD  is  given.      And  the  ratio  of  FD  to  E  is 
given,  wherefore  c  the  ratio  of  GB  to   E  is 
given,  and  AG  is  given  ;  therefore  GB  the  ex- 
cess of  AB  above  a  given  magnitude  AG  has 
a  given  ratio  to  E.  B 

Cor.  1.  And  if  the  first  has  a  given  ratio  to  the  second, 
and  the  excess  of  the  first  above  a  given  magnitude  has  a  given 
ratio  to  the  third ;  the  excess  of  the  second  above  a  given  mag- 
nitude shall  have  a  given  ratio  to  the  third.  For,  if  the  second 
be  called  the  first,  and  the  first  the  second,  this  corollary  will  hv 
the  same  with  the  proposition. 


F- 


D 


a  2.  dat. 

b  19.  5. 
c  9.  dat . 


384  EUCLID'S 

Con.  2.  Also,  if  the  first  has  a  given  ratio  to  the  second,  and 
the  excess  of  the  third  above  a  given  magnitude  has  also  u  given 
ratio  to  the  second,  the  same  excess  shall  have  a  given  ratio  to 
the  first;  as  is  evident  from  the  9th  dat. 


17.  PROP.  XXV. 

IF  there  be  three  magnitudes,  the  excess  of  the 
first  whereof  above  a  gi\'en  magnitude  has  a  given 
ratio  to  the  second ;  and  the  excess  of  the  third  above 
a  given  magnitude  has  a  given  ratio  to  the  same  se- 
cond :  the  first  shall  either  have  a  given  ratio  to  the 
third,  or  the  excess  of  one  of  them  above  a  given 
magnitude  shail  have  a  given  ratio  to  the  other. 

Let  AB,  C,  DE  be  three  magnitudes,  and  let  the  excesses  of 
each  of  the  two  AI?,  DE  above  given  magnitudes  have  given 
ratios  to  C ;  AB,  DE  either  have  a  given  ratio  to  one  another, 
or  the  excess  of  one  of  them  above  a  given  magnitude  has  a  gi- 
ven ratio  to  the  other. 

Let  FB  the  excess  of  AB  above  the  given  magnitude  AF 
have  a  given  ratio  to  C  ;  and  let  GE  the  ex-     A 
cess  of  DE  above  the  given  magnitude  DG 
have  a  given  ratio  to  C  ;  and  because  FB,  GE 
have  each  of  them  a  given  ratio  to  C,  they      F-^ 

a  9.  dat.  have  a  given  ratio  a  to  one  another.  l?ut  to  FB 
GE  the  given  magnitudes  AF,  DG  are  add- 

b  18.  dat.  ed  ;  therefore  b  the  whole  magnitudes  AB, 
DE  have  either  a  given  ratio  to  one  another, 
or  the  excess  of  one  of  them  above  a  given 
magnitude  has  a  given  ratio  to  the  other. 

18.  PROP.  XXVL 

IF  there  be  three  magnitudes,  the  excesses  of  one 
(jf  which  above  given  magnitudes  have  given  ratios 
to  the  other  two  magnitudes ;  tliese  t\\o  shall  either 
have  a  given  ratio  to  one  anoti  er,  or  the  excess  of  one 
of  them  L:bove  a  given  magnitude  sliall  httvc  a  given 
ratio  to  the  other. 


D 

G- 

B 

C 

E 

t)ATA. 


386 


Lfet  AB,  CD,  F.F  be  three  magnitudes,  andlet  GD  the  ex- 
cess of  one  of  them  CO  above  the  given  magnitude  CG  have  a 
given  ratio  to  AB  ;  and  also  let  KD  the  excess  of  the  same  CD 
above  the  given  magnitude  CK  have  a  given  ratio  to  EF,  either 
AB  has  a  given  ratio  to  EF,  or  the  excess  of  one  of  them  above 
a  given  magnitude  has  a  given  ratio  to  the  other. 

Because   GD  has  a  given  ratio  to  AB,  as  GD  to  AB,  so 
make  CG  to  HA  ;  therefore  the  ratio  of  CG  to  HA  is  given  ; 
and  C(i  is  given,  Avherefore  a  HA  is  given  ;  and  because  as  Gi>  *  2.  dat. 
to  AB,  so  is  CG  to  HA,  and  so  is  b  CD  to  HB  ;  the  ratio  of  Cb  b  12. 5. 
to  HB  is  given  :  also  because  Kl)  has  a  given  ratio  to  i..F,  as 
K  J  to  P2F,  so  make  CK  to  LE  ;  therefore  rr 
the  ratio  of  CK  to  LE  is  given  :  and  CK  is 
given,  vi^herefore   LE  a  is  given  :  and  be- 
cause as  KD  to  EF,  so  is  CK  to  L  • ,  and  so  « 
bisCDtoLF;  the  ratio  of  'D  to  LF  is 

given:  but  the  ratio  of  CD  to  HB  is  given  :  ^^  c9.dat, 

wherefore  c  the  ratio  of  HB  to  LF  is  given : 
and  from  HB,  LF  the  given  magnitudes  HA,  B 
LE  being  taken,  the  remainders  AB,  EF 
shall  either  have  a  given  ratio  to  one  another,  or  the  excess  of 
one  of  them  above  a  given  magnitude  has  a  given  ratio  to  the  j  jg  ^^^ 
other  d. 


c 

L 

-  G    - 

K  - 

E   - 

D 

F 

Another  Demonstration. 


Let  AB,  C,  DE  be  three  magnitudes,  and  let  the  excesses  of 
one  of  them  C  above  given  magnitudes  have  given  ratios  to  AB 
and  DE  ;  either  AB,  D2  have  a  given  ratio  to  one  another,  or 
the  excess  of  one  of  them  above  a  given  magnitude  has  a  given 
ratio  to  the  other. 

Because  the  excess  of  C  above  a  given  magnitude  has  a  given  a  14.  dat, 
ratio  to  AB  ;  therefore  a  AB  together  with  a  given  magnitude 
has  a  given  ratio  to  C  :  let  this  given  magni- 
tude be  AF,  wherefore  FB  has  a  given  ratio  to 
C  :  also  because  the  excess  of  C  above  a  given 
magnitude  has  a  given  ratio  to  DE  ;  therefore  a      ' 
DE  together  with  a  given  magnitude   has  a 
given  ratio  to  C  :  let  this  given  magnitude  be       , 
D(t,  wherefore  GE  has  a  given  ratio  to  C :  and 
FB  has  a  given  ratio  to  C,  therefore b  the  ratio  of  FB  to  GE  is 
given  :  and  from  FB,  (iE  the  given  magnitudes  AF,  DG  being 
taken,  the  remainders  AB,  DE  either  have  a  given  ratio  to  one 
another,  or  the  excess  of  one  of  them  above  a  given  magnitude 
has  a  given  ratio  to  the  other  c.  c  19.  dat. 

,3C 


b  9.  dati 


;86 


EUCLID'S 


19. 


PROP.  XXVII. 


a  2.  dat. 


b  19.  5. 


c  9.  dat. 


IF  there  be  three  magnitudes,  the  excess  of  the 
first  of  which  above  a  given  magnitude  has  a  given 
ratio  to  the  second ;  and  the  excess  of  the  second 
above  a  given  magnitude  has  also  a  given  ratio  to  the 
third :  the  excess  of  the  first  above  a  given  magnitude 
shall  have  a  given  ratio  to  the  third. 


Let  AB,  CD,  E  be  three  magnitudes,  the  excess  of  the  first 
of  which  AB  above  the  given  magnitude  AG,  viz.  Ci  B,  has  a 
given  ratio  to  CD  ;  and  FD  the  excess  of  CD  above  the  given 
magnitude  CF,  has  a  given  ratio  to  E  :  the  excess  of  AB  above 
a  given  magnitude  has  a  given  ratio  to  E. 

Because  the   ratio  of  GB  to  CD  is  given,  as  GB  to  CD,  so 
make   GH  to  CF  ;  therefore  the  ratio  of  GH   . 
to  CF  is  given  ;  and  CF  is  given,  w^herefore  a 
GH   is  given;  and  AG   is  given,   wherefore  ^^ 
the  whole  AH  is  given :  and  because  as  GB 
to  CD,  so  is  GH  to  CF,  and  so  is  b   the  re- 
mainder HBto  the  remainder  FD  ;  the  ratio  H-- 
of  HB  to  FD  is  given  :  and  the  ratio  of  FD 
to  E  is  given,  wherefore  c  the  ratio  of  HB  to 
E  is  given  :  and  AH  is  given  ;  therefore  HB  B 
the  excess  of  AB  above  a  given  magnitude  AH  has  a  given  ra 
tio  to  E. 


C 

F- 

D 


d  24.  dat. 


Otherwise.^ 

Let  AB,  C,  D  be  three  magnitudes,  the  excess  EB  of  the 
first  of  which  AB  above  the  given  magnitude  AE  has  a  given 
ratio  to  C,  and  the  excess  of  C  cibove  a  given  . 
magnitude  has  a  given  ratio  to  D  :  the  excess 
of  AB  above  a  given  magnitude  has  a  given  p 
ratio  to  D. 

Because  EB  has  a  given  ratio  to  C,  and  the 
excess  of  C  above  a  given  magnitude  has  a  giv- 
en ratio  to  D  ;  therefore  d  the  excess  of  EB 
above  a  given  magnitude  has  a  given  ratio  to 
D  :  let  this  given  magnitude  be  EF  :  therefore 
FB  the  excess  of  EB  above  EF  has  a  given  ra-  ^  C  D 
tio  to  D  :  and  AF  is  given,  because  AE,  I'.F 


F^ 


DATA. 


387 


are  given  :  therefore  FB  the  excess  of  AB  above  a  given  mag- 
nitude AF  has  a  given  ratio  to  D, 


PROP.  XXVIII. 


25. 


IF  two  lines  given  in  position  cut  one  another, 
the  point  or  points  in  which  they  cut  one  another  ai'e 
given. 

Let  two  lines  AB,  CD  given  in  position  cut  one  another  in 
the  point  E  ;  the  point  E  is  gi-  C 

ven. 

Because  the  lines  AB,  CD 
are  given  in  position,  they  have 
always  the  same  situations,  and 
therefore  the  point,  or  points, 
in  which  they  cut  one  another 
have  always  the  same  situation  : 
and  because  the  lines  AB,  CD  can 
be  found  a,  the  point,  or  points, 
in  which  they  cut  one  another, 
are  likewise  found ;  and  therefore 
are  given  in  position  a. 


See  N. 


a  4.  def. 


PROP.  XXIX. 


26. 


IF  the  extremities  of  a  straight  line  be  given  in  po- 
sition ;  the  straight  line  is  given  in  position  and  mag- 
nitude. 


Because  the  extremities  of  the  straight  lin&  are  given,  they 
can  be  found  a  ;  let  these  be  the  points  A,  B,  between  which  a  4.  def. 
a  straight  line  AB  can  be  drawn b  ;  b  1    Post- 

this  has  an  invariable  position,  be-     A    ■       ■    -B  ulate 

cause    between  two   given    points 

there  can  be  drawn  but  one  straight  line :  and  when  the  straight 
line  AB  is  drawn,  its  magnitude  is  at  the  same  time  exhibited^ 
or  given :  therefore  the  straight  line  AB  is  given  in  positiop 
and  magnitude. 


J8b  EUCLID'S 


gr.  PROP.  XXX, 


IF  one  of  the  extremities  of  a  straight  Hne  given  in 
position  and  magnitude  be  given  j  the  oiher  extremity- 
shall  also  be  given, 

Let  the  point  A  be  given,  to  wit,  one  of  the  extremities  of  a 
straight  line  given  in  magnitude,  and  which  lies  in  the  straight 
line  AC  given  in  position  ;  the  other  extremity  is  also  given. 
Because  the  straight  line  is  given  in  magnitude,  one  equal 

a  1  def.     to  it  can  be  found  a  ;  let  this  be  the  straight  line  D  :  from  the 
greater  straight  line  AC  cut  oft'  AB 
equal  to  the  lesser  D  :  therefore  the     A  B         C 

other  extremity  B  of  the   straight     -— 1 

line  A B  is  found:  and  the  point  B 

has  always  the  same  situation  ;  be-     D 

cause  any  other  point  in  AC,  upon  . 

the  same  side  of  A,  cuts  oft"  between 

it  and  the  point  A  a  greater  or  less  straight  line  than  AB,  that 

b  4  def.  is,  than  D  ;  therefore  the  point  B  is  given  b  :  and  it  is  plain 
another  such  point  can  be  found  in  AC  produced  upon  the  other 
side  of  the  point  A. 


28. 


PROP.  XXXI. 


IF  a  straight  line  be  drawn  through  a  gi^en  point 
parallel  to  a  straight  line  given  in  position;  that  straight 
line  is  given  in  position. 

Let  A  be  a  given  point,  and  BC  a  straight  line  given  in  posi- 
tion ;  the  straight  line  drawn  through  A  parallel  to  BC  is  given 
in  position. 

j^    2    1  Through  A  draw  a  the  straight  line 

DAE    parallel  to   BC ;    the    straight  D  A         E 

line  DAE  has  always  the  same  posi-  ', - 

tion,  because  no  other  straight  line  can 

be  drawn  tl. rough  A  parallel  to  BC;  B  C 

therefore  the  straight  line  DAE  wliich  ■    .    . 

>>  4  def.     has  been  found  is  givenb  in  position. 


DATA.  389 


PROP.  XXXII.  29. 

IF  a  straight  line  be  drawn  to  a  given  point  in  a 
straight  line  given  in  position,  and  makes  a  given  angle 
with  it ;  that  straight  line  is  given  in  position. 

Let  AB  be  a  straight  line  given  in  position,  and  C  a  given 
point  in  it,  the  straight  line  drawn 
to  C,  which  makes  a  given  angle 
with  CB,  is  given  in  position. 

Because  the  angle  is  given,  one 
equal  to  it  can  be  found  a  :  let  this 
be  the  angle  at  D,  at  the  given 
point  C,  in  the  given  straight  line 
AB,  make  bthe  angle  ECB  equal 
to  the  angle  at  D  :  therefore  the 
straight  line  EC  has  always  the 
same  situation,  because  any  other 
straight  line  FC,   drawn  to  the 

point  C,  makes  with  CB  a  greater  or  less  angle  than  the  angle 
ECB,  or  the  angle  at  D  :  therefore  the  straight  line  EC,  which 
has  been  found,  is  given  in  position. 

It  is  to  be  observed,  that  there  are  two  straight  lines  EC,  GC 
upon  one  side  of  AB  that  make  equal  angles  with  it,  and  which 
make  equal  angles  with  it  when  produced  to  the  other  side. 


PROP.  XXXIII. 

IF  a  straight  line  be  dra^^a"l  from  a  given  point  to 
a  straight"  line  given  in  position,  and  makes  a  given 
angle  with  it,  that  straight  line  is  given  in  position. 

From  the  given  point  A,  let  the  straight  line  AD  be  drawn 
to  the  straight  line  ;  C  given  in  position,  and  make  with  it  a 
given  angle  ADC ;  AD  is  given  in  po-  E  A  F 
sition.  ^- 

Through   the   point   A,    draw  a  the  \  a  31.  1. 

straight  line  EAF  parallel  to  BC  ;   and  \ 

because  through  the  given  point  A,  the  „ ^j;^ ~ 

straight  line  EAF  is  draAvn  parallel  to 

BC,  which  is  given  in  position,  EAF  is  therefore  given  in  po- 

sitionb  :  and  because  the  straight  line  AD  meets  the  parallels  b  31.  dat. 


S90 


EUCLID'S 


c  29.  1. 


B<",  EF,  the  angle  EADc  is  equal  to  the  angle  ADC ;  and  ADC 
is  given,  wherefore  also  the  angle  EAD  is  given  :  therefore, 
because  the  straight  line  DA  is  drawn  to  the  given  point  A  in 
tlie  straight  line  EF  given  in  position,  and  makes  with  it  a  given 
d  32.  dat.  angle  EAD,  AD  is  givend  in  position. 


31. 


PROP.  XXXIV. 


See  N.  IF  from  a  given  point  to  a  straight  line  given  in  po- 
sition, a  straight  line  be  drawn  which  is  given  in  mag- 
nitude ;  the  same  is  also  given  in  position. 


B 
D- 


C 


Let  A  be  a  given  point,  and  BC  a  straight  line  given  in  posi- 
tion, a  straight  line  given  in  magnitude  drawn  from  the  point 
A  to  BC  is  given  in  position. 

Because  the  straight  line  is  given  in  magnitude,  one  equal  to 

a  1.  def.  it  can  be  founds ;  let  this  be  the  straight  line  D  :  from  the 
point  A  draw  AE  perpendicular  to  BC  ;  and  /^ 

because  AE  is  the  shortest  of  all  the  straight 
lines  Avhich  can  be  draAvn  from  the  point  A 
to  BC,  the  straight  line  D,  to  which  one 
equal  is  to  be  drawn  from  the  point  A  to 
BC,  cannot  be  less  than  AE.  If  therefore 
D  be  equal  to  AE,  AE  is  the  straight  line 
given  in  magnitude  drawn  from  the  given  point  A  to  BC  :  and 

b  33.  dat.  it  is  evident  that  AE  is  given  in  position b,  because  it  is  drawn 
from  the  given  point  A  to  BC,  which  is  given  in  position,  and 
makes  with  BC  the  given  angle  AEC. 

But  if  the  stredght  line  D  be  not  equal  to  AE,  it  must  be 
greater  than  it :  produce  AE,  and  make  AF  equal  to  D  ;  and 
from  the  centre  A,  at  the  distance  AF,  describe  the  circle  GFH, 
and  join  Ao,  AH  :  because  the  circle  GFH  is  given  in  posi- 

c  6.  dcf.  tionc,  and  the  straight  line  BC  is  also  given  in  position  ;  there- 
fore their  intersection  G  is  gi-  A 

d  28.  dat  yen  d ;  and  the  point  A  is  gi- 
ven ;  wherefore  AG  is  given  in 

e  29.  dat.  position e,  that  is,  the  straight 
line  AG  given  in  magnitude, 
(for  it  is  equal  to  D)  and  drawn 
from  the  given  point  A  to  the 
straight  line  BC  given  in  posi- 
tion, is  also  given  in  position:  and  in  like  manner  AH  is  given, 
in  po.sition  :  therefore  in  this  case  there  are  two  straight  lines 


DATA.  391 

AG,  AH  of  the  same  given  magnitude  which  can  be  drawn  from 
a  given  point  A  to  a  straight  line  BC  given  in  position. 


PROP.  XXXV. 


32, 


IF  a  'Straight  line  be  drawn  between  two  paia'ilel 
straight  lines  given  in  position,  and  makes  given  an- 
gles with  them,  the  sti'aight  hne  is  given  in  magni- 
tude. 

Let  the  straight  line  EF  be  drawn  between  the  parallels  AB, 
CD,  which  are  given  in  position,  and  make  the  given  angles 
BEF,  EFD  :   EF  is  given  in  magnitude. 

In  CD  take  the  given  point  (i,  and  through  G  draw  a  GHa  31.  1. 
parallel  to  EF:  and  because  CD  meets  the  parallels  GH,  EF, 
the  angle  EFD  is  equal  b  to  the  angle  .  t?     tj  tj  b  29.  1. 

HGD  :   and  EFD  is  a  given  angle  ;_ ^     ^  ^ 

wherefore  the  angle  HGD  is  given ;  and 
because  HG  is  drawn  to  the  given  point 
G,inthe  straight  line  CD,  given  in  posi- 
tion, and  makes  a  given  angle  HGD  :  7;  ^1  ~^.  ^ 
the  straight  line  HG  is  given  in  posi- 
tion c :  and  AB  is  given  in  position  :  therefore  the  point  H  is  c  32.  dat. 
given  d  ;  and  the  point  G  is  also  given,  wherefore  GH  is  given  d  28.  <lat 
in  magnitude  e  :  and  EF  is  equal  to  it,  therefore  EF  is  given  ine  29,  dat. 
magnitude. 


PROP.  XXXVI.  23. 

IF  a  straight  line  gi\en  in  magnitude  be  drawn  be-  See  n. 
tween  two  parallel   straight  lines  given  in  pc^itio.i,  it 
shall  make  given  angles  with  the  parallels 

Let  the  straight  line  EF  given  in  magnitude  be  dravm  be-- 
tween  the  parallel  straight  lines  AB,  CD, 
which  are  given  in  position :  the  ar.gles  A  E    H       B 

.\EF,  EFC  shall  be  given.  j      i 

Because  EF  is  given  in  magnitude,  a  i      j 

straight  line  equal  to  it  can  be  found  a  :  j      l  .  i    i  f 

let  this  be  G  :   in  AB  take  a  given    point L_i "    " '  *^  ' 

H,  and  from  it  draw  b  HK  perpendicu-  C               F     K       D  b  12.  1 . 
lar  to  CD  :  therefore  the  straight  line  G,         (\ ^ 


Sd2  EUCLID'S 

that  IS,  EF  cannot  be  less  than  HK  :  and  if  G  be  equal  to  HK, 
EF  also  is  equal  to  it ;  wherefore  EF  is  at  right  angles  to  CD; 
for  if  it  be  not,  EF  would  be  greater  than  HK,  which  is  absurd. 
Therefore  the  angle  EFD  is  a  right,  and  consequently  a  given 
angle. 

But  if  the  straight  line  G  be  not  equal  to  HK,  it  must  be 
greater  than  it :  produce  HK,  and  take  HL,  equal  to  G  ;  _  and 
from  the  centre   H,  at  the  distance  HL,  describe  the  circle 

o  6.  def.     MLN,  and  join  HM,  HN  :  and  because  the  circle  c  MLN,  and 
the  straight  line  CD,  are  given  in  position,  the  points  M,  N  are 

d  28.  dat.    given :  and  the  point  H  is  gi- 
ven, wherefore  the  straight  A  E  H  B 
lines  HM,  HN,  are  given 

e  29.  dat.  in  position  e  :  and  CD  is 
given  in  position ;  therefore 
the  angles  HMN,  HNM, 

f  A.  def.  are  given  in  position  f :  oftbe 
straight  lines  HM,  HN,  let 
HN  be  that  which  is  not  pa- 
rallel to  EF,  for  EF  cannot 


h  29.  1. 


be  parallel  to  both  of  them ;  and  draw  EO  parallel  to  HN :  EO 
S  24-  1-  therefore  is  equal  g  to  HN,  that  is,  to  G ;  and  EF  is  equal  to  G, 
wherefore  EO  is  equal  to  EF,  and  the  angle  EFO  to  the  angle 
EOF,  that  ish,  to  the  given  angle  HNM,  and  because  the  angle 
HNM,  Avhich  is  equal  to  the  angle  EFO,  or  EFD,  has  been 
found;  therefore  the  angle  EFD,  that  is,  the  angle  AEF,  is 
k  1.  def.     given  in  magnitude  k  ;  and  consequently  the  angle  EFC. 

E.  PROP.  XXXVII. 

See  Note.  jjT  ^  straight  line  given  in  magnitude  be  dra\\n  from 
a  point  to  a  straight  line  given  in  posilion,  in  a  gi^-en 
angle;  the  straigTit  line  drawn  thronph  that  point  pa- 
rallel to  the  straight  line  given  in  posiuon,  is  given  in 
position. 

Let  the  straight  line  AD  given  in  magnitude  be  drawn  fron> 
the  point  A  to  the  straight  line  EC  given 

in  position,  in  the  given  angle  ADC :  the  F         AH  F 

straight  line  EAF  drawn  through  A  pa-""^ 
rallel  to  BC  is  given  in  position. 

In  BC  take  a  given  point  G,  and  draw 

Gil  parallel  to  AD  :   and  because  HG  is-; ' — -'; jr, 

drav,n  to  a  given  point  G  in  the  straight "     D     G 


DATA. 


39S 


line  BC  given  in  position,  in  a  given  angle  HGC,  for  it  is  equal  a  29.  1. 
a  to  the  given  angle  ADC  ;  HGis  given  in  position  b  ;  but  it  is  b  32.  dat. 
given  also  in  magnitude,  because  it  is  equal  to  c  AD  which  is  c  34.  1. 
given  in  magnitude  ;  therefore  because  G  one  of  the  extremi- 
ties of  the  straight  line  GH  given  in  position  and  magnitude  is 
given,  the  other  extremity  H  is  given  d ;  and  the  straight  line  d  30.  dat. 
EAF,  which  is  drawn  through  the  given  point  H  parallel  to  bC 
given  in  position,  is  therefore  given  e  in  position.  e  31  dat. 


PROP.  XXXVIII. 

IF  a  straight  line  be  drawn  from  a  given  point  to 
two  parallel  straight  lines  given  in  position,  the  ratio 
of  the  segments  between  the  given  point  and  the  pa- 
rallels shall  be  given. 

Let  the  straight  line  EFG  be  drawn  from  the  given  point  E 
to  the  parallels  AB,  CD,  the  ratio  of  EF  to  EG  is  given. 

From  the  point  E  draw  EHK  perpendicular  to  CD  ;  and 
because  from  a  given  point  E  the  straight  line  EK  is  drawn  to 
CD  which  is  given  in  position,  in  a  given  angle  EKC  ;  EK  is 


.34- 


given  in  position  a  ;  and  AB,  CD  are  given  in  position ;  there»a  33.  dat. 
fore  b  the  points  H,  K  are  given :  and  the  point  E  is  given  ;  b  28.  dat. 
wherefore  c  EH,  EK  are  given  in  magnitude,  and  the  ratio  d  ofc  29.  dat. 
them  is  therefore  given.     But  as  EH  to  EK,  so  is  EF  to  EG,  ,j  j  j^^_ 
because  AB,  CD  are  parallels  ;  therefore  the  ratio  of  EF  to  EG 
is  given. 


PROP.   XXXIX. 


35,36. 


IF  the  ratio  of  the  segments  of  a  straight  line  be-  See  N. 
t\veen  a  given  point  in  it  and   two  parallel  sti-aight 
lines,  be  given,  if  one  of  the  parallels  be  given  in  po- 
sition, the  other  is  also  given  in  position. 

3  D 


894 


EUCLID'S 


From  the  given  point  A,  let  the  straight  line  AED  be  drawn 
to  the  two  parallel  straight  lines  FG,  BC,  and  let  the  ratio  of 
the  segments  AE,  AD  be  given ;  if  one  of  the  parallels  BC  be 
given  in  position,  the  other  FG  is  also  given  in  position. 

From  the  point  A,  draw  AH  perpendicular  to  BC,  and  let 
it  meet  FG  in  K  :  and  because  AH  is  drawn  from  the  given 
point  A  to  the  straight  line  BC  given  in  position,  and  makes  a 


a  33.  dat.  given  angle  AHD  ;  AH  is  given  *  in 
position  ;  and  BC  is  likewise  given  in 
position,  therefore  the  point  H  is  giv-_B 
en  b :  the  point  A  is  also  given ;  where- 
fore AH  is  given  in  magnitude  c,  and, 
because  FG,  BC  are  parallels,  as  AE 
to  AD,   so  is  AK  to  AH  ;    and  the 


b  28.  dat 
c  29.  dat 


E     K 


H 


D 


G 


F 

ratio  of  AE  to  AD  is  given,  where- 

fore  the  ratio  of  AK  to  AH  is  given  ;  but  AH  is  given  in  mag- 
nitude, therefore  d  AK  is  given  in  magnitude  ;  and  it  is  also 
given  in  position,  and  the  point  A  is  given ;  wherefore  e  the  point 
K  is  given.  And  because  the  straight  line  FG  is  drawn  through 
the  given  point  K  parallel  to  EC  which  is  given  in  position, 
f  31.  d.at.    therefore  f  ¥G  is  given  in  position. 


d  2.  dat. 
e  30.  dat 


37.  38. 


PROP.  XL. 


See  N. 


IF  the  ratio  of  the  segments  of  a  straight 


hne  into 
which  it  is  cut  by  three  p:millel  straight  lines,  be  giv- 
en;  iftwoof  the  parallels  are  given  in  position,  the 
position. 


third  also  is  given  ii 


Let  AB,  CD,  HK  be  three  parallel  straight  lines,  of  which 
AB,  CD  are  given  in  position  ;  and  let  the  ratio  of  the  seg- 


DATA. 


895 


ments  GE,  GF  into  which  the  straight  line  GEF  is  cut  by  the 
three  parallels,  be  given  ;  the  third  parallel  HK  is  given  in  po- 
sition. 

In  AB  take  a  given  point  L,  and  draw  LM  perpendicular  to 
CD,  meeting  HK  in  N  ;  because  LM  is  drawn  from  the  given 
point  L  to  CD  which  is  given  in  position  and  makes  a  given 
angle  LMD  ;  LM  is  given  in  positional,  and  CD  is  given  in  ^  33.  dat, 
position,  wherefore  the  point  M  is  given  b  ;  and  the  point  L  isb  28.  dat. 
given,  LM  is  therefore  given  in  magnitude c  ;  and  because  thee  29.  dat. 
ratio  of  GE  to  GF  is  given,  and  as  GE  to  GF,  so  is  NL  to 


K  A 


A 

E 

] 

L 

B 

P 

/" 

V 

K 

/ 

1 

M 


D 


NM ;  the  ratio  of  NL  to  NM  is  given  ;  and  therefore  d  the  ratio 
of  ML  to  LN  is  given ;  but  LM  is  given  in  magnitude  d,  where- d 
force  LN  is  given  in  magnitude ;  and  it  is  also  given  in  position, 
and  the  point  L  is  given;  wherefore  f  the  point  N  is  given,  f 
and  because  the  straight  line  HK  is  drawn  through  the  given 
point  N  parallel  to  CD  which  is  given  in  position,  therefore 
HK  is  given  in  position g,  S' 


r  Cor. 
<.  6.  or 
(  7.dat. 
2.  dat. 
30.  dat. 


31.  dat. 


PROP.  XLL  P. 

IF  a  straight  line  meets  three  parallel  straight  lines  see  Note, 
which  are  gi\'en  in  position,  the  segments  into  which 
they  cut  it  have  a  given  ratio. 

Let  the  parallel  straight  lines  AB,  CD,  EF  given  in  position, 
be  cut  by  the  straight  line  GHK  ;  the  ratio  of  GH  to  HK  is 
given. 

In  AB  take  a  given  point  L,  and  A  G      L  B 

draw  LM  perpendicular  to  CD,  meet- 7 

ing  EF  in  N ;  therefore  a  LM  is  given  , ,  /    ^ 

in  position;   and  ED,  CF  are  given  ^ '-  / 

in  position,  wherefore  the  points  M,  / 

Nare  given;  and  the  point  L  is  given;  / 

therefore  b  the  straight  lines  LM,MN ; 

r^re  gFven  in  magnitude ;  and  the  ratio  i^      "•  -^ 


D 


a  33.  dat, 


b  29.  dat. 


396 


EUCLID'S 


c  1.  dat.    of  LM  to  MN  is  therefore  givenc  :  but  as  LM  to  MN,  so  is 
GH  to  HK ;  wherefore  the  ratio  of  GH  to  HK  is  given. 


39. 


PROP.  XLII. 


See  ifote.      IF  each  of  the  sides  of  a  triangle  be  given  in  mag- 
nitude, tlie  triangle  is  given  in  species. 


a  22.  1. 


b8.  1. 

c  1.  def. 

d  1.  def. 
e  3.  def. 


Let  each  of  the  sides  of  the  triangle  ABC  be  given  in  mag- 
nitude, the  triangle  ABC  is  given  in  species. 

Make  a  triangle  a  DEF,  the  sides  of  which  are  equal,  each 
to  each,  to  the  given  straight  lines  AB,  BC,  CA,  Avhich  can  be 
done  ;  because  any  two  of  them  must  be  greater  than  the 
third ;  and  let  DE  be  e- 
qual  to  AB,  EF  to  BC, 
and  FD  to  CA ;  and  be- 
cause the  two  sides  ED, 
DF  are  equal  to  the  two 
BA,  AC,  each  to  each, 
and  the  base  EF  equal  to  B 
the  base  BC  ;   the   angle 

EDF,  is  equal b  to  the  angle  BAC  ;  therefore,  because  the  an- 
gle EDF,  which  is  equal  to  the  angle  BAC,  has  been  found,  the 
angle  BAC  is  givenc,  in  like  manner  the  angles  at  B,  C  are 
given.  And  because  the  sides  AB,  BC,  CA  are  given,  their 
ratios  to  one  another  are  given d,  therefore  the  triangle  ABC  is 
given  e  in  species. 


40. 


a  23.  1. 


PROP.  XLIIL 

IF  each  of  the  angles  of  a  triangle  be  given  in  mag- 
nitudc,  the  triangle  is  given  in  species. 

Let  each  of  the  angles  of  the  triangle  ABC  be  given  in  mag- 
nitude, the  triangle  ABC  is    given 
in  species. 

Take  a  straight  line  DE  given  in  >1  D 

position  and  magnitude,  and  at  the 
points  D,  E  make  a  the  angle  EDF 
equal   to    the    angle  BAC   and  tlie 

angle  DEF  equal  to  ABC  ;   there-  t>  C       E  I 

fore  the  other  angles  EFD,  BCA  are 
equal,  and  each  of  the  angles  at  the  points  A,  B,  C,  is  given 


DATA.  39/ 

\('herefore  each  of  those  at  the  pomts  D,  E,  F  is  given  :  and 
because  the  straight  line  FD  is  drawn  to  the  given  point  D  in 
DE  which  is  given  in  position,  making  the  given  angle  EDF ; 
therefore  DF  is  given  in  positionb.  In  like  manner  EF  also  is  ^  32.  dat, 
given  in  position ;  wherefore  the  point  F  is  given :  and  the  points 
D,  E  are  given;  therefore  each  of  the  straight  lines  DE,  EF, 
FD  is  given c  in  magnitude  ;  wherefore  the  triangle  DEF  is' 


29.  dat. 


ivhich  is  therefore  given  in  species.  C  4.  6. 

-      1  def. 


given  in  species d  :  and  it  is  similar e  to  the  triangle  ABC  : '^ '*-•  ^^*^^' 

4. 
1 
6. 

PROP.  XLIV.  41. 

IF  one  of  the  angles  of  a  triangle  be  given,  and  if 
tlie  sides  about  it  have  a  given  ratio  to  one  another; 
the  triangle  is  given  in  species. 

Let  the  triangle  ABC  have  one  of  its  angles  BAG  given,  and 
let  the  sides  BA,  AC  about  it  have  a  given  ratio  to  one  another; 
the  triangle  ABC  is  given  in  species. 

Take  a  straight  line  DE  given  in  position  and  magnitude, 
and  at  the  point  D  in  the  given  straight  line  DE,  make  the  an- 
gle EDF  equal  to  the  given  angle  BAC  ;  wherefore  the  angle 
EDF  is  given  ;  and  because  the  straight  line  FD  is  drawn  to 
the  given  point  D  in  ED  which  is  given  in  position,  making 
the  given  angle  EDF;  therefore  FD  A 

is  given  in  position  a.  And  because 
the  ratio  of  BA  to  AC  is  given, 
make  the  ratio  of  ED  to  DF  the 
same  with  it,  and  join  EF  ;  and  be- 


a  32.  dat. 
D 


cause  the  ratio  of  ED  to  DF  is  given,     ^  C         E        F 

and  ED  is  given,  therefore b  DF  is  given  in  magnitude:  and  it^  2.  dat, 

is  given  also  in  position,  and  the  point  D  is  given,  wherefore 

the  point  F  is  given c  ;  and  the  points  D,  E  are  given,  where- ^  30.  dat^ 

fore  DE,  EF,  FD  are  givend  in  magnitude  :  and  the  triangle 'I  29.  dat. 

DEF  is  therefore  given  in  species  ;  and  because  the  triangles e  42.  dat. 

ABC  DEF  have  one  angle  BAC  equal  to  one  angle  EDF,  and 

the  sides  about  these  angles  proportionals  ;  the  triangles  ai"e 

1"  similar  ;  but  the  triangle  DEF  is  given  in  species,  and  there- ^  ^-  ^'• 

fore  also  the  triangle  ABC. 


398 


EUCLID'S 


42.  PROP.  XLV. 

See  Note.      JF  the  sicles  of  a  trlimerle  have  to  one  another  sriven 

is  given  in  species. 


ratios;  the  triangi 


a  2.  dat. 


b  22.  5. 


c  20    1. 
d  A.  5. 


e  22.  1. 

f  42.  dat, 
S  5.  6. 


D     F      F 


Let  the  sides  of  the  triangle  ABC  have  given  ratios  to  one 
another,  the  triangle  ABC  is  given  in  species. 

Take  a  straight  line  D  given  in  magnitude  ;  and  because  the 
ratio  of  AB  to  BC  is  given,  make  the  ratio  of  D  to  E  the  same 
with  it ;  and  D  is  giv«jn,  therefore  a  E  is  given.  And  because 
the  ratio  of  EC  to  C  A  is  given,  to  this  make  the  ratio  of  E  to  F 
the  same  ;  and  E  is  given,  and  therefore  a  F  ;  and  because  as 
AB  to  BC,  so  is  D  to  E  ;  by  composition  AB  and  BC  together 
are  to  BC,  as  D  and  E  to  F  ; 
but  as  BC  to  CA,  so  is  E  to  F ; 
therefore,  f.r  aqualih^  as  AB 
and  BC  are  to  CA,  so  are  D 
and  E  to  F,  and  AB  and  EC 
are  greaterc  than  CA  ;  there- 
fore D  and  E  are  greaterd  than 
F.  In  the  same  manner  any 
two  of  the  three  D,  E,  F  are 
greater  than  the  third.  Makee 
the  triangle  GHK  whose  sides 

are  equal  to  D,  E,  F,  so  that  GH  be  equal  to  D,  HK  to  E,  and 
KG  to  F  ;  and  because  D,  E,  F  are  each  of  them  given,  there- 
fore GH,  HK,  KG  are  each  of  them  given  in  magnitude  ;  there- 
fore the  triangle  GHK  is  givenf  in  species  :  but  as  AB  to  BC, 
so  is  (D  to  E,  that  is)  GH  to  HK  ;  and  as  BC  to  CA,  so  is 
(E  to  F,  that  is,)  HK  to  KG  ;  therefore,  ex  cecjuali,  as  AB 
to  AC,  so  is  GH  to  GK.  Wherefore K^  the  triangle  ABC  is 
equiangular  and  similar  to  the  triangle  GHK ;  and  the  triangle 
GHK  is  given  in  species  ;  therefore  also  the  triangle  ABC  is 
given  in  species. 

Cor  If  a  triangle  is  required  to  be  made,  the  sides  of  which 
shall  have  the  same  ratios  which  three  given  straight  lines  D, 
E,  F  have  to  one  another ;  it  is  necessary  that  every  two  of 
them  be  trreater  than  the  third. 


DATA. 


39^ 


PROP.  XLVI. 


IV  the  Sides  of  a  right-angled  triangle  iibout  one  of 
the  acute  angles  have  a  given  ratio  to  one  anotlier;  the 
triangle  is  given  in  species. 


Let  the  sides  AB,  BC  about  the  acute  angle  ABC  of  the  tri- 
angle ABC,  Avhich  has  a  right  angle  at  A,  have  a  given  ratio  to 
one  another;  the  triangle  ABC  is  given  in  species. 

Take  a  straight  line  DE  given  in  position  and  magnitude  ; 
and  because  the  ratio  of  AB  to  BC  is  given,  make  as  AB  to 
BC,  so  DE  to  EF  ;  and  because  DE  has  a  given  ratio  to  EF, 
and  DE  is  given,  therefore  a  EF  is  given  ;  and  because  as  AB 
to  BC,  so  is  DE  to  EF  ;  and  AB  is  lessb  than  BC,  therefore 
DE  is  lessc  than  EF.  From  the  point  D  draw  DO  at  right  an- 
gles to  DE,  and  from  the  centre 
E  at  the  distance  EF,  describe  a 
circle  which  shall  meet  DG  in 
two  points  ;  let  G  be  either  of 
them,  and  join  EG  ;  therefore 
the  circumference  of  the  circle 
is  given  d  in  position  ;  and  the 

straight  line  DG  is  given e  in  position,  because  it  is  drawn  to 
the  given  point  D  in  DE  given  in  position,  in  a  given  angle  ; 
therefore  f  the  point  G  is  given  ;  and  the  points  D,  E  are  given, 
wherefore  DE,  EG,  GD  are  given  ff  in  magnitude,  and  the  tri- 
angle DEG  in  speciesh.  And  because  the  triangles  ABC,  DEG 
have  the  angle  BAC  equal  to  the  angle  EDG,  and  the  sides 
about  the  angles  ABC,  DEG  proportionals,  and  each  of  the 
other  angles  BCA,  EGD  less  than  a  right  angle  ;  the  triangle 
ABC  is  equiangular!  and  similar  to  the  triangle  DEG  :  but 
DEG  is  given  in  species  ;  therefore  the  triangle  ABC  is  given 
in  species :  and,  in  the  same  manner,  the  triangle  made  by 
drawing  a  straight  line  from  E  to  the  other  point  in  which  the 
circle  meets  DG  is  given  in  species. 


a  2  dat. 
b  19.  1. 
c  A.  5. 


d  6.  def. 
e  32.  dat, 

f  28.  dat, 
g  29.  dat, 
h  42.  dat. 


i  7.  6, 


400 


EUCLID'S 


44, 


PROP.  XLVII. 


See  Note.  IF  a  triancie  has  one  of  its  an  ales  which  is  not  a 
right  angle  given,  and  if  the  sides  about  another  angle 
have  a  given  ratio  to  one  another;  the  triangle  is  given 
in  species. 


Let  the  triangle  ABC  have  one  of  its  angles  ABC  a  given, 

but  not  a  right  angle,  and  let  the  sides  BA,  AC  about  another 

angle  BAC  have  a  given  ratio  to  one  another ;  the  triangle 

ABC  is  given  in  species. 

First,  let  the  given  ratio  be  the  ratio  of 

equality,  that  is,  let  the  sides  B A,  AC  and 

consequently  the  angles  ABC,  ACB  be  e- 

qual ;  and  because  the  angle  ABC  is  given, 
a  32.  1.     the  angle  ACB,  and  also  the  remaining  a 

angle  BAC  is  giveil  ;  therefore  the  trian- 
b  43.  dat.  gle  ABC  is  given  b  in  species  :  and  it  is  B  C 

evident  that  in  this  case  the  given  angle  ABC  must  be  acute. 
Next,  let  the  given  ratio  be  the  ratio  of  a  less  to  a  greater, 

that  is,  let  the  side  AB  adjacent  to  the  given  angle  be  less  than 

the  side  AC  :  take  a  straight  line  DE  given  in  position  and 

magnitude,  and  make  the  angle  DEF  equal  to  the  given  angle 
c  32.  dat.  ABC  ;  therefore  EF  is  given  c  in  position  ;  and  because  the 

ratio  of  B  A  to  AC  is  given,  as  B  A 

to  AC,  so  make  ED  to  DO  ;  and 

because  the  ratio  of  ED  to  DG  is 
and    ED    is     civen,     the 


given, 


D 


d  2.  dat     straight  line  DG  is  given  d,  and 
BA  is  less  than  AC,  therefore  ED 
e  A.  5.     is    lessc    than  DG.      From    the 
centre  D  at  the  distance  DG  de- 
scribe the  circle  GF  meeting  EF 
in  F,  and  join   DP' ;  and  because 
the  circle  is  given f  in  position,  as 
also  the  straight  line  El'\  the  point 
dat.  F  is  given ff  ;  and  the  points  D,  E 
are  given  ;  wherefore  the  straigiit 
'^^^- lines  DE,  EF,  FD  are  givenh  in    ^ 
(].^t    magnitude,  and  the  triangle  DhF 

1.     in  species  i,  and  because  BA  is  less  than  AC,  the  angle  ACB  is 
1  1.  7.  1.   Icssk  than  the  angle  ABC,  and  therefore  ACB  is  less  I  than 


f  6.  def. 


g28. 
h  29. 

i  42 
k  18. 


DATA. 


401 


c  32.  dat. 


a  right  angle.  In  the  same  mannei-,  because  ED  is  less  than 
DG  or  DF,  the  angle  DFE  is  less  than  a  right  angle  :  and  be- 
cause the  triangles  ABC,  DEF  have  the  angle  ABC  equal  to 
the  angle  DEF,  and  the  sides  about  the  angles  BAC  EDF 
proportionals,  and  each  of  the  other  angles  ACB,  DFE  less 
than  a  right  angle  ;  the  triangles  ABC,  DEF  are  m  similar,  and"^  ?"•  ^• 
DEF  is  given  in  species,  wherefore  the  triangle  ABC  is  also  gi- 
ven in  species. 

Thirdly,  Let  the  given  ratio  be  the  ratio  of  a  greater  to  a 
less,  that  is,  let  the  side  AB  adjacent  to  the   given  angle  be  ^ 

greater  than  AC  ;   and  as  in  the  last  A 

case,  take  a  straight  line  DE  given  in 
position  atid  magnitude,  and  make  the 
angle  DEF  equal  to  the  given  angle 
ABC  ;  therefore  FE  is  given  c  in  posi- 
tion :  also  draw  DG  perpendicular  to 
EF ;  therefore  if  the  ratio  of  BA  to 
AC  be  the  same  with  the  ratio  of  ED 
to  the  perpendicular  DG,  the  triangles 
ABC,  DEG  are  similarm,  because  the 
angles  ABC,  DEG  are  equal,  and  DGE 
is  a  right  angle :  therefore  the  angle 
ACB  is  a  right  angle,  and  the  triangle 
ABC  is  given  inb  species. 

But  if,  in  this  last  case,  the  given  ratio  of  BA  to  AC  be 
not  the  same  with  the  ratio  of  ED  to  DG,  that  is,  with  the 
ratio  of  BA  to  the  perpendicular  AM  drawn  from  A  to  BC  ; 
the  ratio  of  BA  to  AC  must  be  less  thano  the  ratio  of  BA 
to  AM,  because  AC  is  greater  than  AM.  Make  as  B  A  to  AC  o  8. 5 
so  ED  to  DH ;   therefore  the  ratio  of  A 

ED  to  DH  is  less  than  the  ratio  of  (BA 
to  AM,  that  is,  than  the  ratio  of)  ED 
to  DG  ;  and  consequently,  DH  is  great- 
erp  than  DG  ;  and  because  B  A  is  great- 
er than  AC,  ED  is  greatere  than  DH. 
From  the  centre  D,  at  the  distance  DH, 
describe  the  circle  KHF  which  necessa- 
rily meets  the  straight  line  EF  in  two 
points,  because  DH  is  greater  than  DG, 
and  less  than  DE.  Let  the  circle  meet 
EF  in  the  points  F,  K  which  are  given, 
as  was  shown  in  the  preceding  case ;  and 
DF,DKbeingjoined,thetrianglesDEF, 

DEK  are  given  in  species,  as  was  there  shown.  From  the  cen- 
tre A,  at  the  distance  AC,  describe  a  circle  meeting  BC  again  in 
L  :  and  if  the  angle  ACB  be  less  than  a  right  angle,  ALM  must 

3E 


b  43.  dat, 


p  10.  5. 
e  A.  5. 


40S 


EUCLID'S 


m  7.  6. 


be  greater  than  a  right  angle  ;  and  on  the  contrary.  In  the  same 
manner,  if  the  angle  DFE  be  less  than  a  right  angle,  DKE  must 
be  greater  than  one  ;  and  on  the  contrary.  Let  each  of  the  an- 
gles ACB,  DEF  be  either  less  or  great-  A 
er  than  a  right  angle  :  and  because  in  the 
triangles  ABC,  DEF  the  angles  ABC, 
DEF  are  equal,  and  the  sides  BA,  AC, 
and  ED,  DF  about  two  of  the  other 
angles  proportionals,  the  triangle  ABC 
is  similar  m  to  the  triangle  DEF.  In  the 
same  manner,  the  triangle  ABL  is  simi- 
lar to  DEK.  And  the  triangles  DEF, 
DEK  are  given  in  species  ;  therefore  al- 
so the  triangles  ABC,  ABL  are  given  in 
species.  And  from  this  it  is  evident,  that 
in  this  third  case,  there  are  always  two 
triangles  of  a  different  species,  to  which 
the  things  mentioned  as  given  in  the  proposition  can  agree. 


45. 


PROP.  XLVIII. 


IF  a  triangle  has  one  angle  given,  and  if  both  the 
bides  together  about  that  angle  have  a  given  ratio  to 
the  remaining  side;  the  triangle  is  given  in  species. 

Let  the  triangle  ABC  have  the  angle  BAC  given,  and  let  the 

sides  BA,  AC  together  about  that  angle  have  a  given  ratio  to 

BC  ;  the  triangle  ABC  is  given  in  species. 

Bisect  a  the  angle  BAC  by  the  straight  line  AD  ;  therefore 

the  angle  BAD  is  given.     And  because  as  BA  to  AC,  so  is  b 

BD  to  CD,  by  permutation,  as  AB  to  BD, 

so  is  AC  to  CD  ;  and  as  BA  and  AC  to- 
gether toBC,  soisc  AB  to  BD.     But  the 

ratio   of  BA   and   AC   together  to  BC  is 

given,  wherefore  the  I'atio  of  AB  to  BD 

is  given,  and  the  angle  BAD  is  given  ; 
d  47.  dat.  therefore  d  the  triangle   ABD  is  given  in 

species,  and  the  angle  ABD  is  therefore  given  ;  the  angle  BAC 

a  43.  dat.  is  also  given,  wherefore  the  triangle  ABC  is  given  in  species  e. 

A  triangle  which  shall  have  the  things  that  are  mentioned 

in  the  proposition  to  be  given,  can  be  found  in  the  folio-wing 


a  9. 
b3, 


cl2. 


•a9.  1. 


DATA.  m 

manner.  Let  EFG  be  the  given  angle,  and  let  the  ratio  of  H 
to  K  be  the  given  ratio  which  the  two  sides  about  the  angle 
EFG  must  have  to  the  third  side  of  the  triangle  ;  therefore 
because  two  sides  of  a  triangle  are  greater  than  the  third  side, 
the  ratio  of  H  to  K  must  be  the  ratio  of  a  greater  to  a  less. 
Bisect  a  the  angle  EFG  by  the  straight  line  FL,  and  by  the ' 
47th  proposition  find  a  triangle  of  which  EFL  is  one  of  the 
angles,  and  in  which  the  ratio  of  the  sides  about  the  angle  op- 
posite to  FL  is  the  same  with  the  ratio  of  H  to  K  :  to  do 
which,  take  FE  given  in  position  and  magnitude,  and  draw  EL 
perpendicular  to  FL  :  then  if  the  ratio  of  H  to  K  be  the  same 
with  the  ratio  of  FE  to  EL,  produce  EL,  and  let  it  meet  FG 
in  P  ;  the  triangle  FEP  is  that  which  was  to  be  found  :  for  it 
has  the  given  angle  EFG  ;  and 
because  this  angle  is  bisected  by 
FL,  the  sides  EF,  FP  together 
are  to  EP,  as  b  FE  to  EL,  that 
is,  as  H  to  K. 

But  if  the  ratio  of  H  to  K 
be  not  the  same  with  the  ratio 
of  FE  to  EL,  it  must  be  less  than  N 

it,  as  was  sho'wn  in  prop.  47,  and  in  this  case  their  are  two  tri- 
angles, each  of  which  has  the  given  angle  EFL,  and  the  ratio  of 
the  sides  about  the  angle  opposite  to  FL  the  same  with  the  ratio 
of  H  to  K.  By  prop.  47,  find  these  triangles  EFM,  EFN,  each 
of  which  has  the  angle  EFL  for  one  of  its  angles,  and  the  ratio 
of  the  side  FE  to  EM  or  EN  the  same  with  the  ratioof  H  to  K ; 
and  let  the  angle  EMF  be  greater,  and  ENF  less  than  a  right 
angle.  Artd  because  H  is  greater  than  K,  EF  is  greater  than 
EL,  and  therefore  the  angle  EFN,  that  is,  the  angle  NFG,  is 
less  f  than  the  angle  ENF.  To  each  of  these  add  the  angles  f  18.  1. 
NEF,  EFN  :  therefore  the  angles  NEF,  EFG  are  less  than  the 
angles  NEF,  EFN,  FNE,  that  is,  than  two  right  angles ;  there- 
fore the  straight  lines  EN,  FG  must  meet  together  when  produ- 
ced ;  let  them  meet  in  O,  and  produce  EM  to  G.  Each  of  the 
triangles,  EFG,  EFO  has  the  things  mentioned  to  be  given  in 
the  proposition  :  for  each  of  them  has  the  given  angle  EFG  ; 
and  because  this  angle  is  bisected  by  the  straight  line  FMN,  the 
sides  FF,  FG  together  have  to  EG  the  third  side  the  ratio  of 
FE  to  EM,  that  is,  of  H  to  K.  In  like  manner,  the  sides  EF, 
FO  together  have  to  EO  the  ratio  which  H  has  to  K. 


404  EUCLID'S 

46.  PROP.  XLIX. 

IF  a  triangle  has  one  angle  gi^"e^,  and  if  the  sides 
about  another  angle,  both  together,  have  a  given  ra- 
tio to  the  third  side  ;  the  triangle  is  given  in  species. 

Let  the  triangle  ABC  have  one  angle  ABC  given,  and  let  the 
two  sides  B A,  AC  about  another  angle  BAC  have  a  given  ratio 
to  BC  ;  the  triangle  ABC  is  given  in  species. 

Suppose  the  angle  BAC  to  be  bisected  by  the  straight  line 
AD  ;  BA  and  AC  together  are  to  BC,  as  AB  to  BD,  as  was 
shown  in  the  preceding  proposition.  But  the  ratio  of  BA  and 
AC  together  to  BC  is  given,  therefore  also  the  ratio  of  AB  to 
a  44.  dit.  BD  is  given.  And  the  angle  ABD  is  given,  wherefore  a  the 
triangle  ABD  is  given  in  species  :  and  consequently  the  angle 
B.AD,  and  its  double  the  angle  BAC  A 

are  given  ;  and  the  angle  ABC  is  gi- 
ven.    Therefore  the  triangle  ABC  is 
b  4".  dat.   given  in  species b. 

A  triangle  which  shall  have  the 
things  mentioned  in  the  proposition 
to  be  given,  may  be  thus  found.  Let 
EFG  be  the  given  angle,  and  the  ra- 
tio of  H  to  K  the  given  ratio  ;  and 
by  prop.  44,  find  the  triangle  EFL, 
which  has  the  angle  EFG  for  one  of 
its  angles,  and  the  ratio  of  the  sides 
EF,  FL  about  this  angle  the  same  with 
the  ratio  of  H  to  K ;  and  make  the  angle  LEM  equal  to  the 
angle  FEL.  And  because  the  ratio  of  H  toK  is  the  ratio  which 
two  sides  of  a  triangle  have  to  the  third,  H  must  be  greater  than 
K  ;  and  because  EF  is  to  FL,  as  H  to  K,  therefore  EF  is  great- 
er than  FL,  and  the  angle  FEL,  that  is,  LEM,  is  therefore  less 
than  the  angle  ELF.  Wherefore  the  angles  LFE,  FEM  are 
less  than  two  right  angles, as  was  sho^vn  in  the  foregoing  propo- 
sition, and  the  straight  lines  FL,  EM  must  meet  if  produced  ; 
let  them  meet  in  G,  EFG  is  the  triangle  which  was  to  be  found  ; 
for  EFG  is  one  of  its  angles,  and  because  the  angle  FFG  is  bi- 
sected by  EL,  the  two  sides  FE  EG  together  have  to  the  third 
side  FG  the  ratio  of  EF  to  FL  that  is,  the  given  ratio  of  H  to  K. 


DATA.  40S 


PROP.  L. 


76. 


IF  from  the   vertex  of  a  triangle  gi\en  in  species, 
a  sti-aigbt  line  be  dra%\Ti  to  the  base  in  a  gi^  en  angle ; 

it  shall  liave  a  gi\"en  ratio  to  the  base. 

From  the  vertex  A  of  the  triangle  ABC  which  is  given  in 
species,  let  AD  be  drawn  to  the  base  BC  in  a  given  angle  ADB : 
the  ratio  of  AD  to  BC  is  given. 

Because  the  triangle  ABC  is  given  in  spe- 
cies, tlie  angle  ABD  is  given,  and  the  angle 
ADB  is  given,  therefore  the  triangle  ABD  is 

given  a  in  species ;  wherefore  the  ratio  of  AD     ^^^        I      \   i  43.  dat. 
to  AB  is  given.     And  the  ratio  of  AB  to  BC 
is  given  ;  and  therefore  b  the  ratio  of  AD  to 
BC  is  given. 


PROP.  LI.  47. 

RECTILINEAL  figures  given  in  species,  are  di- 
^  ided  into  triangles  whicli  are  given  in  species. 

Let  the   rectiligeal  figure   .ABCDE   be  given  in  species ' 
ABCDE  may  be  dixided  into  triangles  given  in  species. 

Join  BE,  BD ;  and  because  ABCDE  is  given  in  species,  the 
angle  BAE  is  given  a,  and  the  ratio  of  BA  j^^  a  3.  de£, 

to  AE  is  given  a  ;  wherefore  the  triangle 
BAE  is  given  in  species  b,  and  the  angle 
AEB  is  therefore  given  a.  But  the  whole 
angle  AED  is  given,  and  therefore  the  re-  B 
maining  angle  BED  is  given,  and  the  ratio 
of  AEto  EB  is  given,  as  also  the  ratio  of 
AE  to  ED  ;  therefore  the  ratio  of  BE  to         C  D 

ED  is  given  c.  And  the  angle  BED  is  given,  wherefore  the  tri-  c  9.  dat 
angle  BED  is  given  b  in  spjecies.     In  the  same  manner,  the  tri- 
angle BDC  is  given  in  species:  therefore  rectilineal  figures 
which  aix^  given  in  species  are  divided  into  triangles  given  in 
species. 


406  EUCLID'S 

48.  PROP.  LII. 

IF  two  triangles  given  in  species  be  described  upon 
the  same  straight  line ;  they  shall  have  a  given  ratio 
to  one  another. 

Let  the  ti'iangle  ABC,  ABD  given  in  species  be  described 
upon  the  same  straight  line  AB ;  the  ratio  of  the  triangle  ABC 
to  the  triangle  ABD  is  given. 

Through  the  point  C,  draw  CE  parallel  to  AB,  and  let  it 
meet  DA  produced  in  E,  and  join  BE.  Because  the  triangle 
ABC  is  given  in  species,  the  angle  BAC,  that  is,  the  angle 
ACE,  is  given  ;  and  because  the  triangle  ABD  is  given  in  spe- 
cies, the  angle  DAB, 
that  is,  the  angle  AEC  E  C 

is  given.  Therefore  the  ^--^  ~      ~]7]    L H 

triangle  ACE  is  given 
ill  species ;  wherefore 
the  ratio  of  E  A  to  AC 
a  3,  def.     is  given  a,  and  the  ra- 
tio of   CA  to  AB  is 
given,  as  also  the  ratio 
of  B  A  to  AD  ;  there- 
b  9.  dat.     fore  the  ratio  of  b  EA  to  AD  is  given,  and  the  triangle  ACB  is 
c  37.  1.      equal  c  to  the  triangle  AEB,  and  as  the  triangle  AEB,  or  ACB, 
^  J  g  '      is  to  the  triangle  ADB,  so  is  d  the  straight  line  EA  to  AD.  But 
the  ratio  of  EA  to  AD  is  given,  therefore  the  ratio  of  the  trian- 
gle ACB  to  the  triangle  ADB  is  given. 

PROBLEM. 

To  find  the  ratio  of  two  triangles  ABC,  ABD  given  in  species, 
and  which  are  described  upon  the  same  straight  line  AB. 

Take  a  straight  line  FG  given  in  position  and  magnitude, 
and  because  the  angles  of  the  triangles  ABC,  ABD  are  given, 
at  the  points  F,  G  of  the  straight  line  FG,  make  the  angles 
e  23.  1.  GFH,  GFK  e  equal  to  the  angles  BAC,  BAD  ;  and  the  angles 
FGH,  FGK  equal  to  the  angles  ABC,  ABD,  each  to  each. 
Therefore  the  triangles  ABC,  ABD  are  equiangular  to  the  tri- 
angles FGH,  FGK,  each  to  each.  Through  the  point  H  draw 
HL  parallel  to  FG  meeting  KF  produced  in  L.  And  because 
the  angles  BAC,  BAD  are  equal  to  the  angles  GFH,  GFK,  each 
to  each  ;  therefore  the  angles  ACE,  AEC  are  equal  to  FHL 
FLH,  each  to  each,  and  the  triangle  AEC  equiangular  to  the 
triangle  FLH.     Therefore  as  EA  to  AC,  so  is  LF  to  FH  ;  and. 


^       DATA.  40r 

ti 
as  CA  to  AB,  so  HF  to  FG  ;  and  as  BA  to  AD,  so  is  GF  to 

FK ;  whei-efore,  ex  aguali,  as  EA  to  AD,  so  is  LF  to  FK.  But, 
as  was  shown,  the  triangle  ABC  is  to  the  triangle  ABD,  as  the 
straight  line  EA  to  AD,  that  is,  as  LF  to  FK.  The  ratio  there- 
fore of  LF  to  FK  has  been  found,  which  is  the  same  with  the 
ratio  of  t}ie  ti'iangle  ABC  to  the  triangle  ABD. 

PROP.  LIIL  49. 

IF  two  rectilineal  figures  given  in  species  be  de-  ^^^  Note. 
scribed  upon  the  same  straight  hne ;  they  shall  have  a 
given  j'atio  to  one  another. 

Let  any  two  rectilineal  figures  ABCDE,  ABFG  Avhich  are 
given  in  species,  be  described  upon  the  same  straight  line  AB; 
the  ratio  of  them  to  one  another  is  given. 

Join  AC,  AD,  AF;  each  of  the  triangles  AED,  ADC,  ACB, 
AGF,  ABF  is  given  a  in  species.     And  becausethetrianglesa51.dat. 
ADE,  ADC  given  in  species  are  de- 
scribed upon  the   same  straight  line 

AD,  the  ratio  of  EAD  to  DAC  is  E   .  ,  ,  ^  b    2  1 

given  b ;    and,    by  composition,   the        \       /      ^^^"'y'        b52.  ca. 
ratio  of  EACD  to  DAC  is  given c        \   X^^-^       /„     c7.dat. 
And  the  ratio  of  DAC  to  CAB  is 
given  b,  because  they  are  described 
upon  the   same    straight  line   AC  ; 
therefore  the  ratio  of  EACD  to  ACB  ^^  ^ 

is  given d  ;  and  by  composition,  the      jj ^y .y j_j__o    ^  ^'  *^**' 

ratio  of  ABCDE  to  ABC  is  given. 

In  the  same  manner,  the  ratio  of  ABFG  to  ABF  is  given.  But 
the  ratio  of  the  triangle  ABC  to  the  triangle  ABF  is  given  ; 
wherefore b,  because  the  ratio  of  ABCDE  to  ABC  is  given,  as 
also  the  ratio  of  ABC  to  ABF,  and  the  ratio  ABF  to  ABFG  ; 
the  ratio  of  the  rectilineal  ABCDE  to  the  rectilineal  ABFG  is 
given  d. 

PROBLEM. 

To  find  the  ratio  of  two  rectilineal  figures  given  in  species, 
and  described  upon  the  same  straight  line. 

Let  ABCDE,  ABFG  be  two  rectilineal  figures  given  in 
species,  and  described  upon  the  same  straight  line  AB,  and 
join  AC,  AD,  AF.  Take  a  straight  line  HK  given  in  positioi\ 
and  magnitude,  and  by  the  52d  dat.  find  the  ratio  of  the  tri- 
angle ADE  to  the  triangle  ADC,  and  make  'che  ratio  of  HK 


408 


EUCLID'S 


to  KL  the  same  with  it.  Find  also  the  ratio  of  the  triangle  ACD 
to  the  triangle  ACB.  And  make  the  ratio  of  KL  to  LM  the 
same.  Also,  find  the  ratio  of  the  triangle  ABC  to  the  triangle 
ABF,  and  make  the  ratio  of  LM  to  MN  the  same.  And  lastly, 
find  the  ratio  of  the  triangle  AFB  to  the  triangle  AFG,  and 
make  the  ratio  of  MN  to  NO  the  D 

same.  Then  the  ratio  of  ABCDE 
to  ABFG  is  the  same  with  the  ra- 
tio of  HM  to  MO. 

Because  the  triangle  EAD  is  to 
the  triangle  DAC,  as  the  straight 
line  HK  to  KL ;  and  as  the  triangle 
DAC  to  CAB,  so  is  the  straight 
line  KL  to  LM ;  therefore,  by  using 
composition  as  often  as  the  number 
of  triangles  requires,  the  rectilineal 
ABCDE  is  to  the  triangle  ABC,  as  the  straight  line  HM  to  ML. 
In  like  manner,  because  the  triangle  G  AF  is  to  FAB,  as  ON  to 
NM,  by  composition,  the  rectilineal  ABFG  is  to  the  triangle 
ABF,  as  MO  to  NM;  and,  by  inversion,  as  ABF  to  ABFG,  so 
is  NM  to  MO.  And  the  triangle  ABC  is  to  ABF,  as  LM  to 
MN.  Wherefore,  because  as  ABCDE  to  ABC,  so  is  HM  to 
ML  ;  and  as  ABC  to  ABF,  so  is  LM  to  MN  ;  and  as  ABF  to 
ABFG,  so  is  MN  to  MG  ;  ex  xquali^  as  the  rectilineal  ABCDE 
to  ABFG,  so  is  the  straight  line  HM  to  MO; 


50.  PROP.  LIV. 

IF  two  straight  lines  have  a  gi\  en  ratio  to  one  ano- 
ther; the  similar  rectilineal  figin-es  described  upon  them« 
similarly,  shall  have  a  given  ratio  to  one  another. 

Let  the  straight  lines  AB,  CD  have  a  given  ratio  to  one  ano- 
ther, and  let  the  similar  and  similarly  placed  rectilineal  figures 
E,  F  be  described  upon  them  ;  the  ratio  of  E  to  F  is  given. 

To  AB,  CD,  let  G  be  a  third  pro' 
portional :  therefore  as  AB  to  CD, 
so  is  CD  to  G.  And  the  ratio  of  AB 
to  CD  is  given,  wherefore  the  ratio 
of  CD  to  G  is  given  ;  and  conse- 
quently the  ratio  of  AB  to  G  is  also 

a  9.  dat.     given  a  .     But  as  AB  to  G,  so  is  the 

b  2.  Cor.  figure  E  to  the  figure  ^F.     There- 

20.  6.        fore  the  ratio  of  E  to  F  is  given. 


DATA.  409 

PROBLEM. 

To  find  the  ratio  of  two  similar  rectilineal  figures,  E,  F,  simi- 
larly described  upon  straight  lines  AB,  CD  which  have  a  given 
ratio  to  one  another :  let  G  be  a  third  proportional  to  AB,  CD. 

Take  a  straight  line  H  given  in  magnitude ;  and  because  the 
ratio  of  AB  to  CD  is  given,  make  the  ratio  of  H  to  K  the  same 
with  it ;  and  because  H  is  given,  K  is  given.  As  H  is  to  K,  so 
make  K  to  L  ;  then  the  ratio  of  E  to  F  is  the  same  with  the 
ratio  of  H  to  L  :  for  AB  is  to  CD,  as  H  to  K,  wherefore  CD  is 
to  G,  as  K  to  L ;  and,  ex  xquali,  as  AB  to  G,  so  is  H  to  L :  but 
the  figure  E  is  tob  the  figure  F,  as  AB  to  G,  that  is,  as  H  to  L.  ^^  ^°^' 

PROP.  LV.  51 

IF  two  straight  lines  have  a  given  ratio  to  one  ano- 
ther ;  the  rectihneal  figures  given  in  species  described 
upon  them,  shall  have  to  one  anodier  a  given  ratio. 

Let  AB,  CD  be  two  straight  lines  which  have  a  given  ratio 
to  one  another  ;  the  rectilineal  figures  E,  F  given  in  species 
and  described  upon  them,  have  a  given  ratio  to  one  another. 

Upon  the  straight  line  AB,  describe  the  figure  AG  similar 
and  similarly  placed  to  the  figure  F  ;  and  because  F  is  given  in 
species,  AG  is  also  given  in  spe-  ^V 

cies :  therefore,  since  the  figures  /   xtK 

E,  AG  which  are  given  in  spe-     a  /-^ -S  B     C         D 

cies,  are  described  upon  the  same  I 1 

straight  line  AB,  the ,  ratio  of  E  [     ^     j 

to  AG  is  given  a,  and  because  the  a  53.  dat. 

ratio  of  AB  to  CD  is  given,  and     jj K L— — 

upon  them  are  described  the  simi- 
lar and  similarly  placed  rectilineal  figures  AG,  F,  the  ratio  of 
AG  to  F  is  given  b  :  and  the  ratio  of  AG  to  E  is  given  ;  there-  b  54.  dat. 
fore  the  ratio  of  E  to  F  is  givenc.  ^  g  j^j.^ 

PROBLEM. 

To  find  the  ratio  of  two  rectilineal  figures  E,  F  given  in  spe- 
cies and  described  upon  the  straight  lines  AB,  CD  which  have 
a  given  ratio  to  one  another. 

Take  a  straight  line  H  given  in  magnitude  ;  and  because 
the  rectilineal  figures  E,  AG  given  in  species  are  described  up- 
on the  same  straight  line  AB,  find  their  I'atio  by  the  53d  dat, 
and  make  the  ratio  of  H  to  K  the  same  ;  K  is  thex'efore  given  : 
and  because  the  similar  rectilineal  figures  AG,  F  are  described 

3  F     ' 


410 


EUCLID'S 


62. 


a  53. 


b  2.  dat 


14.  5. 


53. 


upon  the  straight  lines  AB,  CD,  which  have  a  given  ratio,  find 
their  ratio  by  the  54th  dat.  and  make  the  ratio  of  K  to  L  the 
same  :  the  figure  E  has  to  F  the  same  ratio  which  H  has  to  L : 
for,  by  the  construction,  as  E  is  to  AG,  so  is  H  to  K  ;  and  as 
AG  to  F,  so  is  K  to  L  ;  therefore,  ejc  ayuali,  as  E  to  F,  so  is 
H  to  L. 

PROP.  LVI. 

IF  a  rectilineal  figure  given  in  species  be  described 
upon  a  straight  line  given  in  magnitude ;  the  figure  is 
given  in  magnitude. 

Let  the  rectilineal  figure  ABCDE  given  in  species  be  de- 
scribed upon  the  straight  line  AB  given  in  magnitude  ;  the 
figure  ABCDE  is  given  in  magnitude. 

Upon  AB  let  the  square  AF  be  described  ;  therefore  AF  is 
given  in  species  and  magnitude,  and  because  the  rectilineal 
figures  ABCDE,  AF  given  in  species  are 
described  upon  the  same  straight  line  AB, 
dat- the  ratio  of  ABCDE  to  AF  is  given  a: 
but  the  square  AF  is  given  in  magnitude, 
therefore ''  also  the  figure  ABCDE  is 
given  in  magnitude. 

PROB. 

To  find  the  magnitude  of  a  rectilineal  ^ 
figure  given  in  species  described  upon  a 
straight  line  given  in  magnitude. 

Take  the  straight  line  GH  equal  to  the 
given  straight  line  AB,  and  by  the  53d 
dat.  find  the  ratio  which  the  square  AF 
upon  AB  has  to  the    figure    ABCDE  ; 

and  make  the  ratio  of  GH  to  HK  the  same  ;  and  upon  GH  de- 
scribe the  square  GL,  and  complete  the  parallelogram  LHKM ; 
the  figure  ABCDE  is  equal  to  LHKM  :  because  AF  is  to 
ABCDE,  as  the  straight  line  GH  to  HK,  that  is,  as  the  figure 
GL  to  HM  ;  arid  AF  is  equal  to  GL  ;  therefore  ABCDE  is 
equal  to  HMc. 

PROP.  LVH. 

IF  two  rectilineal  figu'-es  are  given  in  species,  and 
if  a  side  of  one  of  tliem  has  a  given  ratio  to  a  side  of 
the  other;  the  ratios  of  the  remaining  sides  to  tlie  re- 
m.aining  sides  shall  be  gi^-en. 


DATA. 


411 


Let  AC,  DF  be  two  rectilineal  figures  given  in  species,  and 
let  the  ratio  of  the  side  AB  to  the  side  DE  be  given,  the  ratios 
of  the  remaining  sides  to  the  remaining  sides  are  also  given.       a  3.  def. 

Because  the  ratio  of  AB  to  DE  is  given,  as  also  a  the  ratios  of  ^  jq  ^^^^ 
AB  to  BC,  and  of  DE  to  EF,  the  ratio  of  BC  to  EF  is  givenb. 
In  the  same  manner,  the  I'atios  of  the 
other   sides   to   the   other  sides  are 
given.  A^ 

The  ratio  Avhich  BC  has  to  EF 
may  be  found  thus :  take  a  straight 
line  G  given  in  magnitude,  and  be- 
cause the  ratio  of  BC  to  BA  is  given, 
make  the  ratio  of  G  to  H  the  same ; 
and  because  the  ratio  of  AB  to  DE  is 
given,  make  the  ratio  of  H  to  K  the 
same  ;  and  make  the  ratio  of  K  to  L 
the  same  with  the  given  ratio  of  DE 

to  EF.  Since  therefore  as  BC  to  BA,  so  is  G  to  H  ;  and  as  BA 
to  DE,  so  is  H  to  K  :  and  as  DE  to  EF,  so  is  K  to  L  :  ejc  aquali, 
BC  is  to  EF,  as  G  to  L  ;  therefore  the  ratio  of  G  to  L  has  been 
found,  which  is  the  same  with  the  ratio  of  BC  to  EF. 


PROP.  LVIII. 


IF  t^vo  similar  rectilineal  figures  ha'\'e  a  given  ratio  See  N. 
to  one  another,  their  homologous  sides  have  also  a 
G:iven  ratio  to  one  another. 

Let  the  two  similar  rectilineal  figures  A,  B  have  a  given  ratio 
ro  one  another,  their  homologous  sides  have  also  a  given  ratio. 

Let  the  side  CD  be  homologous  to  EF,  and  to  CD,  EF  let^  g.  Cor. 
the  straight  line  Gbe  a  third  proportional.     As  therefore  a  CD  20.  6, 
to  G,  so  is  the  figure  A  to  B  ;  and 
the  ratio  of  A  to  B  is  given,  there- 
fore the  ratio  of  CD  to  G  is  given ; 
and  CD,  EF,  G  are  proportionals; 
wherefore  b  the  ratio  of  CD  to  EF 
is  given. 

The  ratio  of  CD  to  EF  may  be 
found  thus  :  take  a  sti'aight  line  H 
given  in  magnitude  ;  and  because 

the  ratio  of  the  figure  A  to  B  is  given,  make  the  ratio  of  H  to  K 
the  same  with  it :  and  as  the  13th  dat.  directs  to  be  done,  find  a 
mean  proportional  L  between  H  and  K  ;  the  ratio  of  CD  to  EF 


l±2  - 

E     F       G 


b  13.  dat 


H 


K 


412  EUCLID'S 

is  the  same  with  that  of  H  to  L.  Let  G  be  a  third  proportional 
to  CD,  EF  ;  therefore  as  CD  to  G,  so  is  (A  to  B,  and  so  is)  H 
to  K  ;  and  as  CD  to  EF,  so  is  H  to  L,  as  is  shown  in  the  13th 
dat. 


54.  PROP.  LIX. 

^eeN.  IF  two  rectilineal  figures  given  in  species  have  a 

given  ratio  to  one  another,  their  sides  shall  likewise 
have  given  ratios  to  one  another. 

Let  the  two  rectilineal  figures  A,  B  given  In  species,  have  a 
given  ratio  to  one  another,  their  sides  shall  also  have  given  ratios 
to  one  another. 

If  the  figure  A  be  similar  to  B,  their  homologous  sides  shall 
have  a  given  ratio  to  one  another,  by  the  preceding  proposition  ; 
and  because  the  figures  are  given  in  species,  the  sides  of  each  of 

a  3.  def.     them  have  given  ratios  a  to  one  another  ;  therefore  each  side  of 

b  9.  dat.     oiie  of  them  has  b  to  each  side  of  the  other  a  given  ratio. 

But  if  the  figure  A  be  not  similar  to  B,  let  CD,  EF  be  any 
two  of  their  sides  ;  and  upon  EF'  conceive  the  figure  EG  to  be 
described  similar  and  similarly 
placed  to  the  figure  A,  so  that 
CD,  EF  be  homologous  sides  ; 
therefore  EG  is  given  in  spe- 
cies ;  and  the  figure  B  is  given 

c  S3,  dat.  iri  species  ;    wherefore  c  the  ra- 
tio of  B  to  EG  is  given  ;  and 

the  ratio  of   A  to  B  is  given,     H 

therefore  b  the  ratio  of  the  figure     K 

A  to  EG  is  given  ;  and  A  is  si-     M 

d  58.  dat.   rnilar  to  EG  ;  therefore  d  the  ra-     L ■ 

tin  of  the  side  CD  to  EF  is  given ;  and  consequently  b  the  ratios 
of  the  remaining  sides  to  the  remaining  sides  are  given. 

The  ratio  of  CD  to  EF  may  be  found  thus  :  take  a  straight 
line  H  given  in  magnitude,  and  because  the  ratio  of  the  figure 
A  to  B  is  given,  make  the  ratio  of  H  to  K  the  same  with  it. 
And  by  the  53d  dat.  find  the  ratio  of  the  figure  B  to  EG,  and 
nnake  the  ratio  of  K  to  L  the  same :  between  H  and  L 
find  a  mean  proportional  M,  the  ratio  of  CD  to  EF  is  the 
same  with  the  ratio  of  II  to  M  ;  because  the  figure  A  is  to  B 
as  II  to  K  ;  and  as  B  to  EG,  so  is  K  to  L  ;  ex  ie(^ua/i,  as  A 


DATA. 


413 


to  EG,  so  is  H  to  L  :  and  the  figures  A,  EG  are  similar,  and 
M  is  a  mean  proportional  between  H  and  L  ;  therefore,  as  wsls 
shown  in  the  preceding  proposition,  CD  is  to  EF  as  H  to  M. 

PROP.  LX. 


55. 


IF  a  rectilineal  figure  be  given  in  species  and  mag- 
nitude, the  sides  of  it  sliall  be  given  in  magnitude. 

Let  the  rectilineal  figure  A  be  given  in  species  and  magnitude, 
its  sides  are  given  in  magnitude. 

Take  a  straight  line  BC  given  in  position  and  magnitude, 
and  upon  BC  describe  a  the  figure  D  similar,  and  similarly  a  IS.  6. 
placed,  to  the  figure  A,  and 
let  EF  be  the  side  of  the 
figure  A  homologous  to 
BC  the  side  of  D  ;  there- 
fore the  figure  D  is  given 
in  species.  And  because 
upon  the  given  straight 
line  BC  the  figure  D  gi- 
ven in  species  is  described, 

D  is  given  b  in  magnitude,  H      |       *lCf  Z' *     b  36.  dat. 

and    the   figure   A    is    gi- 
ven in  magnitude,  therefore 

the  ratio  of  A  to  D  is  given  :  and  the  figure  A  is  similar  to 
D ;  therefore  the  ratio  of  the  side  EF  to  the  homologous  side     ^n 
BC  is  given  c  ;  and    BC  is  given,  wherefore  d  EF  is  given: 
and  the  ratio  of  EF  to  EG  is  given e,  therefore  EG  is  given. '^  2-  '^^*^- 
And,  in  the  same  manner,  each  of  the  other  sides  of  the  figure  ^  3*  '^'cf 
A  can  be  shown  to  be  given. 

PROBLEM. 

To  describe  a  rectilineal  figure  A  similar  to  a  given  figure  D 
and  equal  to  another  given  figure  H.  It  is  prop.  25,  b,  6,  Elem. 

Because  each  of  the  figures  D,  H  is  given,  their  ratio  is  gi- 
ven, which  may  be  found  by  making  f  upon  the  given  straight  f  Ccr.  45. 
line  BC  the  parallelogram  BK  equal  to  D,  and  upon  its  side.. 
CK  making  f  the  parallelogram  KL  equal  to  H  in  the  angle 
KCL  equal  to  the  angle  MBC  ;  therefore  the  ratio  of  D  to  H, 
that  is,  of  BK  to  KL,  is  the  same  with  the  ratio  of  BC  to  CL  : 
and  because  the  figux'es  D,  A  are  similar,  and  that  the  ratio  of 
D  to  A,  or  H,  is  the  same  with  the  ratio  of  BC  to  CL ;  by 
the  58th  dat.  the  ratio  of  the  homologous  sides  BC,  EF  is  the 
same  with  the  ratio  of  BC  to  the  mean  proportional  between 
BC  and  CL.     Find  EF  the  mean  proportional ;  then  EF  is  the 


'414  EUCLID'S 

side  of  the  figure  to  be  described,  homologous  to  BC  the  side 

of  D,  and  the  --gure  itself  can  be  described  by  the   18th  prop. 

g  2.  Cor.   book  6,  which,  by  the  construction,  is  similar  to  D ;  and  because 

20.  6.         D  is  to  A,  as  g  BC  to  CL,  that  is  as  the  figure  BK  to  KL;  and 

h  14.  5.      ^^^"^  ^  ^^  equal  to  BK,  therefore  A  h  is  equal  to  KL,  that  is,  to  H. 

5^  PROP,  LXL 

See  N.  ^^  ^  parallelogram  given  in  magnitude  lias  one  of 

its  sides  and  one  of  its  angles  given  in  magnitude, 
the  other  side  also  is  given. 

Let  the  parallelogram  ABDC  given  in  magnitude,  have  the 
side  AB  and  the  angle  B  AC  given  in  magnitude,  the  other  side 
AC  is  given. 

Take  a  straight  line  EF  given  in  position  and  magnitude  ; 

and  because  the  parallelogram  AD  A  B 

is  given  in  magnitude,  a  rectilineal  J  7 

^    ,  r.     figure  equal  to  it   can  be  found  ^  •  /  / 

And  a  parallelogram   equal   to  this  /_ / 

h  Cor   45.  ^8'"''^  ^^^  t»e  applied  b  to  the  given  c  D 

straight  line  EF  in  an  angle  equal  to  E  F 

the  given  angle  BAC.     Let  this  be  f 

the    parallelogram     EFHG     having  / 

the  angle   I'EG  equal  to  the  angle         / 

BAC.       And  because  the  parallelo-        [ 

grams  AD,  EH  are  equal,  and  have         G 
the  angles  at  A  and  E  equal ;    the 
c  14  6.       sides  about  them  are  reciprocally  proportional  c  ;   therefore  as 
AB  to  EF,  so  is  EG  to  AC  ;  and  AB,  EF,  EG  are  given,  there- 
d  12.  6.      fore  also  AC  is  given  d.     Whence  the  way  of  finding  AC  is  ma- 
nifest. 
"•  PROP.  LXIL 

See  N.  IF  a  p?.rallelograni  has  a  given  angle,  the  rectangle 

contained  by  the  sides  about  that  angle  has  a  i^ivQu 
ratio  to  the  pcU'alielogram. 

A 
Let  the  parallelogram  ABCD  have  the  gi- 
ven angle  ABC,  the  rectangle  AB,  BC  has 
a  given  ratio  to  the  parallelogram  AC. 

From  the  point  A  draw  AE  perpendi-  B~E 
cular  to  BC  ;  because  the  angle  ABC  is  gi- 
ven, as  also  the  angle  AEB,  the  triangle 
a  43.  dat.  ABE  is  given  a  in  species  ;  therefore  the 
ratio  of  BA  to  AE  is  given.  But  as  BA  to 
b  1.  6.  AE,  so  is  b  t!ie  rectangle  AB,  BC  to  the 
rectangle    AE,  BC;    therefore  the  ratio  of 


&' 


DATA.  415 

the  l-ectangle  AB,  BC  to  AE,  BC  that  is  c,  to  the  parallelogram  c  35. 1. 
AC  is  given. 

And  it  is  evident  how  the  ratio  of  the  rectangle  to  the  pa- 
rallelogram may  be  found  by  making  the  angle  FGH  equal  to 
the  given  angle  ABC,  and  draAving,  from  any  point  F  in  one  of 
its  sides,  FK  perpendicular  to  the  other  GH  ;  for  GF  is  to  FK, 
as  B  A  to  AE,  that  is,  as  the  rectangle  AB,  BC,  to  the  parallelo- 
gram AC. 

Cor.  And  if  a  triangle  ABC  has  a  given  angle  ABC,  the  rect- 
angle AB,  BC  contained  by  the  sides  about  that  angle,  shall 
have  a  given  ratio  to  the  triangle  ABC. 

Complete  the  parallelogram  ABCD  ;  therefore,  by  this  pro- 
position, the  rectangle  AB,  BC  has  a  given  ratio  to  the  paral- 
lelogram AC  ;  and  AC  has  a  given  ratLo  to  its  half  the  triangle  d  d  41.  1. 
ABC  ;  therefore  the  rectangle  AB,  BC  has  a  given e  ratio  to  thee  9.  dat. 
triangle  ABC. 

And  the  ratio  of  the  rectangle  to  the  triangle  is  found  thus  : 
make  the  triangle  FGK,  as  was  shown  in  the  proposition  ;  the 
ratio  of  GF  to  the  half  of  the  perpendicular  FK  is  the  same  with 
the  ratio  of  the  rectangle  AB,  BC  to  the  triangle  ABC.  Because, 
as  was  shown,  GF  is  to  FK,  as  AB,  BC  to  the  parallelogram 
AC  ;  and  FK  is  to  its  half,  as  AC  is  to  its  half,  which  is  the  tri- 
angle ABC  ;  therefore,  ex  equali^  GF  is  to  the  half  of  FK,  as 
AB,  BC  rectangle  is  to  the  triangle  ABC. 


PROP.LXIIL  5^. 

IF  two  parallelograms  be  equiangular,  as  a  side  of 
the  first  to  a  side  of  the  second,  so  is  the  other  side 
of  the  second  to  the  straight  line  to  which  tlie  other 
side  of  the  first  has  the  same  ratio  ^vhich  the  first  pa- 
rallelogram has  to  the  second.  And  consequently, 
if  t'lc  ratio  of  the  first  parallelogram  to  the  second  be 
given,  the  ratio  of  the  other  side  of  the  first  to  that 
straight  line  is  given;  and  if  the  ratio  of  tlie  other 
side  of  tlie  first  to  that  straight  line  be  given,  the  ratio 
of  t!-ic  fii-st  parallelogram  to  the  second  is  given. 

Let  iVC,  DF  be  two  equiangular  parallelograms,  as  BC,  a 
side  of  the  first,  is  to  EF,  a  side  of  the  second,  so  is  DE,  the 
o-tlier  side  of  the  second,  to  the  straiglit  line  to  which  AB,  the 


416  EUCLID'S 

other  side  of  the  first  has  the  same  ratio  wliich  AC  has  to  DF. 

Produce  the  straight  line  AB,  and  make  as  BC  to  EF,  so 
DE  to  BG,  and  complete  the  parallelo- 
gram BGHC  ;  therefore,  because  BC  or 
GH,  is  to  EF,  as  DE  to  BG,  the  sides 
about  the  equal  angles  BGH,  DEF  are 
&  14.  6.  reciprocally  proportional  ;  wherefore  a 
the  parallelogram  BH  is  equal  to  DF  ; 
and  AB  is  to  BG,  as  the  parallelogram 
AC  is  to  BH  ,  that  is,  to  DF  ;  as  there- 
fore BC  is  to  EF,  so  is  DE  to  BG,  Avhich 
is  the  straight  line  to  which  AB  has  the 
same  ratio  that  AC  has  to  DF. 

And  if  the  ratio  of  the  parallelogram  AC  to  DF  be  given,  then 
the  I'atio  of  the  straight  line  AB  to  BGis  given  ;  and  if  the  ratio 
of  AB  to  the  straight  line  BG  be  given,  the  ratio  of  the  paral- 
lelogram AC  to  DF  is  given. 


74.  73.  PROP.  LXIV. 

See  Note.  IF  two  parallelograms  have  unequal  but  given  an- 
gles, and  if  as  a  side  of  the  first  to  a  side  of  the  se- 
cond, so  the  other  side  of  the  second  be  made  to  a 
certain  straight  line ;  if  the  ratio  of  the  first  parallelo- 
gram to  the  second  be  given,  the  ratio  of  the  other 
side  of  the  first  to  that  straight  line  shall  be  given. 
And  if  the  ratio  of  the  other  side  of  the  first  to  that 
straight  line  be  given,  the  ratio  of  the  first  parallelo- 
gram to  the  second  shall  be  given. 

Let  ABCD,  EFGH  be  two  parallelograms  which  have  the 
unequal,  but  given,  angles  ABC,  EFG  ;  and  as  BC  to  FG,  so 
make  EF  to  the  straight  line  M.  If  the  ratio  the  parallelo- 
gram AC  to  EG  be  given,  the  ratio  of  AB  to  M  is  given. 

At  the  point  B  of  the  straiglit  line  BC  make  the  angle 
CBK  equal  to  the  angle  EFG,  and  complete  the  parallelogram 
KBCL.  And  because  the  ratio  of  AC  to  EG  is  given,  and  that 
a  35. 1  AC  is  equal  a  to  the  parallelogram  KC,  therefore  the  ratio  of 
KC  to  EG  is  given;  and  KC,  EG  ai'-a  equiangular;  there- 
fa  63.  dat.  fore  as  BC  to  FG,  so  is  ^  EF  to  the  straight  line  to  which  KB 
has  a  given  ratio,  viz.  the  same  which  the  parallelogram 
KC  has  to  EG  ;  but  as  BC  to  FG,  so  is  EF  to  the  straight 
line  M  ;  therefore  KB  has  a  given  ratio  to  M  ;  and  the  ratio 


DATA. 


•♦ir 


34"  y 


C  b  63.  dat. 


.K 


G 


©f  AB  to  BK  is  given,  because   the  triangle  ABIC  is  given  in 
species  <= ;  therefore  the  ratio  of  AB  to  M  is  given  d.  c  43.  dav. 

And  if  the  ratio  of  AB   to   M  be  ^iven,  the  ratio  of  the  pa-^^^'^i*** 
ralielogram  AC  to  EG  is  given  :  for  since  the  ratio  of  KB  to 
BA  is  given,  as  also  the  ratio  of  AB  to  M,      K     A  L     D 

the  ratio  of  KB  to  M  is  given  ^ ;  and  be- 
cause the  parallelograms  KC,  EG  are 
equiangular,  as  BC  to  FG,  so  is  ^  EF  to 
the  straight  line  to  which  KB  has  the 
same  ratio  which  the  parallelogram  KC 
has  to  EG  ;  but  as  BC  to  FG,  so  is  EF  to 
M  ;  therefore  KB  is  to  M,  as  the  parallel- 
ogram KC  is  to  EG  ;  and  the  ratio  of  KB 

to  M  is  given,  therefore  the  ratio  of  the  parallelogram  KC,  that 
is,  of  AC  to  EG,  is  given. 

Cor.  And  if  two  triangles  ABC,  EFG,  have  two  equal  angles, 
or  two  unequal,  but  given  angles  ABC,  EFG,  and  if  as  BC  a  side 
of  the  first  to  FG  a  side  of  the  second,  so  the  other  side  of  the  se- 
cond EF  be  made  to  a  straight  line  M  ;  if  the  ratio  of  the  trian- 
gles be  given,  the  ratio  of  the  other  side  of  the  first  to  the  straight 
line  M  is  given. 

Complete  the  parallelograms  ABCD,  EFGH  ;  and  because  the 
ratio  of  the  triangle  ABC  to  the  triangle  EFG  is  given,  the  ratio 
of  tile  parallelogram  AC  to  EG  is  given  e,  because  the  parallelo-e  15.  3. 
grams  are  doub4e  f  of  the  triangles  ;  and  because  BC  istoFG,  as  £41.  1. 
EF  to  M,  the  ratio  of  AB  to  M  is  given  by  the  63d  dat.  if  the  an- 
gles ABC,  EFG  are  equal  ;  but  if  they  be  unequal,  but  given  an- 
gles^ the  ratio  of  AB  to  M  is  given  by  this  proposition. 

And  if  the  ratio  of  AB  to  M  be  given,  the  ratio  of  the  paral- 
lelogram AC  to  EG  is  given  by  the  same  proposition  ;  and  there- 
fore the  ratio  of  the  triangle  ABC  to  EFG  is  given. 


73. 


PROP.  LXV. 


^Sf 


If  two  equiangular  parallelograms  have  a  given  ratio 
to  one  another,  and  ii  one  side  has  to  one  side  a  given 
ratio  ;  the  other  side  shall  also  have  to  the  other  side  a 


gl^•en  ratio. 


Let  the  two  equiangular  parallelograms  AB,  CD  have  a  given 
ratio  to  one  another,  and  let  the  side  EB  have  a  given  ratio  to  the 
side  FD  ;  the  other  side  AE  has  also  a  given  ratio  to  the  other 
side  CF, 

3G 


o. 


418  EUCLID'S 

Because  the  two  equiangular  parallelograms  AB,  CD  have  a 
given  ratio  to  one  another  ;  as  EB,  a  side  of  the  first,  is  to  FD, 

a  63.dat.  a  side  of  the  second,  so  is  ^  FC,  the  other  side  of  the  second,  to 
the  straight  line  to  which  AL,  the  other  side  of  the  first,  has 
the  same  given  ratio  \yhich  the  first  parallelogram  AB  has 
to  the  other  CD.  Let  this  straight  line  be  EG;  therefore  the 
ratio  of  AE  to  EG  is  given ; 
and  EB  is  to  FD,  as  FC  to 
EG,  therefore  the  ratio  of  FC  -^j 
to  EG  is  given,  because  the 
ratio  of  EB  to  FD  is  given  ;  E^ 
and  because  the  ratio  of  AE 
to  EG,  as  also  the  ratio  of  FC  G' 
to    EG  is   given  ;    the    ratio    of 

b9,  dat.   AE  to  CF  is  given  i^. 

The  ratio  of  AE  to  CF  may  be  found  thus  :  take  a  straight 
line  H  given  in  magnitude  ;  and  because  the  ratio  of  the  paral- 
lelogram AB  to  CD  is  given,  make  the  ratio  of  H  to  K  the  same 
with  it.  And  because  the  ratio  of  FD  to  EB  is  given,  make 
the  ratio  of  K  to  L  the  same  :  the  ratio  of  AE  to  CF  is  the  same 
with  the  ratio  of  H  to  L.  Make  as  EB  to  FD,  so  FC  to  EG, 
therefore,  by  inversion,  as  FD  to  EB,  so  is  EG  toFC  ;  and  as  AE 
to  EG,  so  is  a  (ilie  parallelogram  AB  to  CD,  and  so  is)  H  to  K  ; 
but  as  LG  to  FC,  so  is  (FD  to  EB,  and  so  is)  K  to  L  ;  therefore, 
ex  aequali,  as  AE  to  FC,  so  is  H  to  L. 


69.  PROP.  LXVI. 


IF  two  parallelogiaiiis  have  unequal,  but  given  an- 
p'les,  and  a  criven  ratio  to  one  another ;  if  one  side  has 
to  one  side  a  giA  en  ratio,  the  other  side  has  also  a  given 
ratio  to  the  other  side. 

Let  the  two  parallelograms  ABCD,  EFGH  which  have  the 
given  unequal  angles  ABC,  EFG,  have  a  given  ratio  to  one  an- 
other, and  let  the  ratio  of  BC  to  EG  be  given  ;  the  ratio  also  of 
AB  to  EF  is  given. 

At  the  point  B  of  the  straight  line  BG  make  the  angle  CBK 
equal    to    the    given    angle    EFG,   and  complete  the  parallelo- 
gram BKLC  ;  and  because  each  of  the  angles  BAK,  AKB  is 
-43  d      S'^'^"'  ^'^^    triangle    ABK.    is    given  ^  in    species  ;  therefore  the 
'ratio    of  AB  to  Bl^  is  given  ;  ynd  because,  by  the  hypothesis; 


DATA. 


A19 


the  ratio  of  the  parallelogram  AC  to  EG  is  given,  and  that  AC 

is  equal  ^   to  BL  ;  therefore  the  ratio  of  BL  to  EG  is  given  :  and  b  35. 1. 

because    BL  is  equiangular  to  EG,  and,  by  the   hypothesis,  the 

ratio  of  BC  to  FG  is  given  ;  therefore  *=  the  ratio  of  kb  to  EFisc65.clat. 

given,  and   the    ratio    of   KB  to  BA  is 

given;     the  ratio    therefore 'i   of  AB  to  \     ,         "  ,        d  9.  dau 

EF  is  given. 

The  ratio  of  AB  to  EF  may  be  found 
thus :  take  the  straight  line  MN  given  E 
in  position  and  magnitude  ;  and    make 
the  angle   NMO    equal    to  the    given  F 
angle  BAK,  and  the  angle  MNO  equal 
to  the  given  angle  EFG  or  AKB  :  and 

because  the  parallelogram  BL  is  equiangular  to  EG,  and  has  a 
given  ratio  to  it,  and  that  the  ratio  of  BC  to  FG  is  given  ;  find 
by  the  65th  dat.  the  ratio  of  KB  to  LF ;  and  make  the  ratio  of 
NO  to  OP  the  same  with  it:  then  the  ratio  of  AB  to  EF  is  the 
same  with  the  ratio  of  MO  to  OP  :  for  since  the  triangle  ABK 
is  equiangular  to  MON,  as  AB  to  BK,  so  is  MO  to  ON :  and  as 
KB  to  EF,  so  is  NO  to  OP  ;  therefore,  ex  squali,  as  AB  to  EF, 
so  is  MO  to  OP. 


PROP.  LXVII. 

IF  the  sides  of  two  equiangular  parallelograms  have  See  x. 
given    ratios  to  one  another ;    the  pai-allelograms  sliall 
have  a  given  ratio  to  one  another. 


Let  ABCD,  EFGH  be  two  equiangular  parallelograms,  and  let 
the  ratio  of  AB  to  EF,  as  also  the  ratio  of  BC  to  F(j,  be  given  ; 
the  ratio  of  the  parallelogram  AC  to  EG  is  given. 

Take  a  straight  line  K  given  in  magnitude,  and  because  the 
ratio  of  AB  to  EF  is  given,  make  A  D     E  II 

the  ratio  of  K  to  L  the  same  with    . 
it;    therefore  L  is   given*:  and    \ 
because  the   ratio  of  BC  to  FG   BV. 
is  given,  make  the  ratio  of  L  to  K 

P.I  the    same  :     therefore    M  is  L  F 

given  a  :  and  K  is  given,  where-  M  — . 

lore^  the  ratio  of  K  to  Mis  given  :    but  the  parallelogram  AC  isbl.  dat. 
to  the  parallelogram  EG,  as  the  straight  line  K  to  the  straigh.t 


a  2.  dat. 


426 


EUCLID'S 


line  M,  as  is  demonstrated  in  the  2^d  prop,  of  B.  6,  Elein.  there- 
fore the  ratio  of  AC  to  EG  is  given. 

From  this  it  is   plain    how  the  ratio  of  two  equiangular  paral- 
lelograms may  be  found  when  the  ratios  of  their  sides  are  given. 


70. 


PROP.  LXVIir, 


SceN. 


IF  the  sides  of  two  parallelograms  \^•llich  have  un- 
equal, but  given  angles,  have  given  ratios  to  one 
another  ;  the  parallelograms  shall  have  a  given  ratio  to 
one  another. 


Let  tvl'o  parallelograms  ABCD,  EFGH  which  have  the  given 
unequal  angles  ABC,  EFG  have  ihe  ratios  of  their  sides,  viz.  of 
AB  to  EF,  and  of  BC  to  FG,  given  ;  the  ratio  of  the  parallelo- 
gram AC  to  EG  is  given. 

At  the  point  B  of  the  straight  line  BC  make  the  angle  CBK 
equal  to  the  given  angle  EFG,  and  complete  the  parallelogram 
KBCL  :   and   because   each  of  the  angles  B.VK,  BK  A,  is    given, 

a  43.  dat.  the  triangle    ABK  is  given  ^  in  species:  therefore  the   ratio   of 
AB  to  BK  is  given  ;  and  the  ratio  of  AB  to  EF  is  given,  where- 

h  9.  dat.  foi-eb  the  ratio   of  BK  to   EF  is  j^  ^  L 

given  :  and  the  ratio  of  BC  to 
FG  is  given  ;  andtlie  angle  KBC 
is    equal    to    the    angle    El'G  ; 

c  67.  dat.  therefore  c  the  ratio  of  the  pa- 
rallelogram KC  to  EG  is  given; 

d.  35.  1.  hut  KC  is  equal  "^  to  AC;  there- 
fore the  ratio  of  AC  to  EG  is 
given. 

The  ratio  of  the  parallelogram  AC  to  EG  may  be  found 
tiius  :  take  the  straight  line  MN  given  in  position  and  magni- 
tude, and  make  the  angle  MNO  equal  to  the  given  angle  KAB» 
and  the  angle  NMO  equal  to  the  given  angle  AKB  or  FEli  : 
and  because  the  ratio  of  AB  to  EF  is  given,  make  the  ratio  of 
NO  to  P  the  same  :  also  make  the  ratio  of  P  to  Q  the  same  with 
the  given  ratio  of  BC  to  FG,  the  parallelogram  AC  is  to  EG,  as 
rvIO  to  Q. 

Because  the  angle  KAB  is  equal  to  the  angle  MNO,  and 
the  angle  AKB  rquul  to  the  angle  N.MO;  the  triangle  AKB  is 
equiangular  to  MiVIO:  therefore  as  KB  to  BA,  so  is  MO  to 
ON;  and  as  BA  to  KV,  so  is  NO  to  P;  wherefore,  ex  a)- 
<juali,  as  KB  to  EF,  so  is  MO  to  P  :    and  BC  is  to  FG,    as  P 


'  DATA.  '4&i 

to  Q,  and  the  parallelograms  KC,  EG  are  equiangular ;  there- 
fore, as  was  shown  in  prop.  67,  the  parallelogram  KC,  that  is, 
AC,  is  to  EG,  as  MO  to  Q.  71. 

Cor.   1.  If  two  triangles  ABC,  DEF  have  two  equal  angles^ 
or  two  unequal,  but  given  angles,  ABC,  DEF,  and  if  the  ratios 
of  the  sides  about  these  angles,  viz.          A             G  D          H 
the  ratios  of  AB  to  DE,  and   of  BC 
to  EF  be  given  ;  the  triangles   shall 
have  a  given  ratio  to  one  another.  

Complete  the  parallelograms  BG,      jj  C         E         F 

EH  :  the  ratio  of  BG  to  EH  is  gi- 
ven a  ;    and  therefore  the   triangles  which   are    the   halves ''  of  a  67.  or 
them  have  a  given  •=  ratio  to  one  another.  ^^'  '^*'^- 

Cor.  2.  If  the  bases  BC,  EF  of  two  triangles  ABC,  DEF  have  ^  "'*•  ^• 
a  given  ratio  to  one  another,  and  if  also  the  straight  lines  AG,  *^  15.5. 
DH  which  are    drawn  to   the    bases  from  the  opposite  angles,      72. 
either  in  equal    angles,   or  unequal,   but    given    angles    AGC) 
DIIF  have   a  given  ratio  to  one     K     A  L     D 

another  :  the  triangles  shall  have 


a  given  ratio  to  one  another. 

DrawBK,  EL  parallel  to  AG, 
DH  and   complete   the    paralle-     !>      G         C 
lograms  KC,  LF.     And  because 

the  angles  AGC,  DHF,  or  their  equals,  the  angles  KBC,  LEF 
are  either  equal,  or  unequal,  but  given  ;  and  that  the  ratio  of  AG 
to  DH,  that  is,  of  KB  to  LE,  is  given,  as  also  the  ratio  of  BC  to 
EF  ;  therefore  ^   the   I'atio  of  the  partvlleiogram  KC  to  LF  isgi-a  67.  or 
ven  ;  wherefore    also  the  ratio  of  the  triangle  ABC  to  DEF  is^^*^^^ 
given' 


C41.1. 
i.15,5. 


PROP.  LXIX.  61. 


IF  a  parallelogram  which  has  a  given  angle  be  ap- 
plied to  one  side  of  a  rectilineal  figure  given  in  species  ; 
if  the  figure  have  a  given  ratio  to  the  parallelogram,  the 
parallelogram  is  given  in  species. 

Let  ABCD  be  a  rectilineal  figure  given  in  species,  and  to  one 
side  of  it  AB,  let  the  parallelogram  ABEF,  having  the  given 
angle  ABE,  be  applied  ;  if  the  figure  ABCD  has  a  given  ratio 
•to  the  parallelogram  BF,  the  p;vraileIogram  BF  is  given  in 
species. 


423 


EUCLID'S 


Throuf^h  the  point  A  draw  AG  parallel  to  BC,  and  through 
the  point  C   draw  CG  parallel  to  AB,  and  produce  GA,  CB  to 
aS.def.   the  points  H,  K:  because  the    angle    ABC    is   given  =>,  and  the 
ratio  of  AB  to   BC  is  given,  the    figure  ABCD  being  given  in 
species;  therefore,  the  parallelogram  BG  is  given ^   in    species. 
And  because  upon  the  same  straight  line  AB  the  two  rectilineal 
figures  BD,  BG  given  in  species  are  described,  the  ratio  of  BD 
.b53.dat.  to  BG  is   given  b;  and,  by    hypothesis,  the    ratio   of  BD  to  the 
e  9.dat.    parallelogram  BF  is  given  ;  wherefore '^  the  ratio  of  BF,  that  is^, 
d35.1.    of  the  parallelogram  BH,  to  BG  is  given,  and  therefore  e  the  ra- 
-e  1.  6.      tio  of  the  straight  line  KB  to  BC  is  given ;  and  the  ratio  of  BC 
to  B  A  is  given,  wherefore  the  ratio  of  KB  to  BA  is  given  "^  :  and 
because  the  angle  ABC  is  given,  the  adjacent  angle  ABK  is  gi- 
ven ;  and  the  angle  ABE  is  given,  therefore  the  remaining  angle 
KBE  is  given.     The  angle  EKB  is  also  given,  because  it  is  equal 
to  the  angle  ABK  ;  therefore  the  triangle  BKE  is  given  in  spe- 
cies, and  consequently  the  ratio  of  EB  to  BK  is  given ;  and  the 
ratio  of  KB  to  B  \  is  given, 
wherefore  "=  the    ratio   of  EB 
to  B A  is  given  ;   and  the  an- 
gle   ABE   is    given,    there- 
fore   the   parallelogram    BF 
is  given  in  species. 

A  parallelogram  similar 
to  BF  may  be  found  thusipi" 
take  a  straight  line  LM  gi- 
ven in  position  and  magnitude  ;  and  because  the  angles  ABK» 
ABE  are  given,  make  the  angle  NLM  equal  to  ABK,  and  the 
angle  NLO  equal  to  ABF..  And  because  the  ratio  of  BF'  to  BD 
is  given,  make  the  ratio  of  LM  to  P  the  same  with  it ;  and  be- 
cause the  ratio  of  the  figure  BD  to  BG  is  given,  find  this  ratio  by 
the  53d  dat.  and  make  the  ratio  of  P  to  Q  the  same.  Also,  be- 
cause the  ratio  of  CB  to  BA  is  given,  make  the  ratio  of  Q  to  R 
the  same  ;  and  take  LN  equal  to  R  ;  through  the  pomi  M  draw 
OM  parallel  to  LN,  and  complete  the  parallelogram  NLOS;  then 
/       this  is  similar  to  the  parallelogram   BF. 

Because  the  angle  ABK  is  equal  to  NLM,  and  the  angle  ABE 
to  NLO,  the  angle  KBE  is  equal  to  MLO  ;  and  the  angles 
EKE,  LMO  are  equal,  because  the  angle  ABK  is  equal  to 
NLlNI ;  therefore  the  triangles  BKE,  LMO  are  equiangular  to 
one  another  ;  wherefore  as  BE  to  BK,  so  is  LO  to  LM  ;  and 
because  as  the  figure  BF  to  BD,  so  is  the  straight  line  LM 
to  P  ;  and  as  BD  to  BG,  so  is  P  to  Q  ;  ex  xquali,  as  BF, 
that  is  'Miirl  to  BG,  so  is  LM  to  Q:  but  BB  is  to^  BG,  as 
KB  to  BC  ;  as  therefore  KB  to  BC,  so  is  LM  to  Q  ;  and  be- 
cause BE  hi  to  BK,  as  LO  to  LM;  and  as  BK  to  BC,  so  is 
LM  to  Q  :    and  as  BC  to  BA,  so  Q  was  made  to  R;    there- 


DATA.  43» 

fore,  ex  aquali,  as  BE  to  BA,  so  is  LO  to  R,  that  is  to  LN  ; 
and  the  angles  ABE,  NLO  arc  equal  ;  therefore  the  parallelo- 
gram BF  is  similar  to  LS. 


PROP.  LXX.  62,  ra 


IF  two  straight  lines  have  a  given  ratio  to  one  ano-SeeNote, 
thcr,  and  upon  one  of  them  be  described  a  rectihiieal 
figure  given  in  species,  and  upon  the  other  a  paraiklo- 
gram  having  a  given  angle ;  if  the  figure  have  a  given 
ratio  to  the  paralielogram,  tlie  paralleiogrLm  is  given  in 
species. 

Let  the  two  straight  lines  AB,  CD  have  a  given  ratio  to  one 
another,  and  upon  AB  let  the  figure  AEB  given  in  species  be 
described,  and  upon  CD  the  parallelogram  DF  having  the  given 
angle  FCD  ;  if  the  ratio  of  AEB  to  DF  be  given,  the  parallelo- 
gram DF  is  given  in  species. 

Upon  the  straight  line  AB,  conceive  the  parallelogram   AG  to 
be  described  similar,  and  similarly  placed  to  FD  ;    and  because 
the  ratio  of  AB  to  CD  is  given,   and  upon  them  are  described 
the  similar  rectiline.il  figures  AG, 
FD  ;  the  ratio  of  AG  to  FD  is  gi- 
ven »;  and  the  ratio  of  FD  to  AEB 
is  given  ;    therefore  ^   the  ratio  of 
AEB  to  AG  is  given;  and  the  angle 
ABG  is   given,  because  it  is  equul 
to  the  angle  FCD  ;   because  tht^re- 
fore  the  parallelogram   AG  which 
has  a  given  angle  ABG  is  applied 
to  a  side  AB  of  the  figure  AEB  t:,i- 
ven  in  species,  and  the  ratio  of  AEB  to  AG  is  given,  the  paral-G69.dat* 
lelogram  AG  is  given  <=  in  species  ;    but  FD  is  similar  to  AG  ; 
therefore  FD  is  given  in  species. 

A  parallelogram  similar  to  FD  may  be  found  thus:  take  a 
straight  line  H  given  in  magnitude  ;  and  because  the  ratio  of  the 
figure  AEB  to  FD  is  given,  make  the  ratio  of  H  to  K  the  same 
with  it :  also,  because  the  ratio  of  the  straight  line  CD  to  AB 
is  given,  find  by  the  54th  dat.  the  ratio  which  the  figure  FD  de- 
scribed upon  CD  has  to  the  figure  AG  described  upon  AB  si- 
milar to  FD  ;  and  make  the  ratio  of  K  to  L  the  same  with  this 
ratio  :   and  because  the  ratios  of  H  to  K,   and  of  K  to  L  are 


4a4r  EUCLID'^S 

b  9.  dat.  given,  the  ratio  of  H  to  L  is  given  ^  ;  because,  therefore,  ns- 
AEB  to  FD,  so  is  H  to  K  ;  and  as  FD  to  AG,  so  is  K  lo  L  ; 
ex  xquali,  a«  AEB  to  AG,  so  is  H  to  L ;  therefore  the  ratio  of 
AEB  to  AG,  is  given  ;  and  the  figure  AEB  is  given  in  spe- 
cies, and  to  its  side  AB  the  parallelogram  AG  is  applied  in  the 
given  angle  AEG  ;  therefore  by  the  69th  dat.  a  paralklogrann 
may  be  found  similar  to  AG  :  let  this  be  the  puraMelogram 
MN ;  MN  also  is  similar  to  FD  ;  for,  by  the  construction,  MN 
is  similar  to  AG,  and  AG  is  similar  to  FD  j  therefore  the  paral 
lelogram  FD  is  similar  to  MN. 


«^  PROP.  Lxxr. 

IF  the  extremes  of  three  proportional  straight  lines- 
have  given  ratios  to  the  extremes  of  other  three  propor- 
tional straight  lines ;  the  means  shall  also  have  a  given 
ratio  to  one  another :  and  if  one  extreme  has  a  given 
ratio  to  one  extreme,  and  the  mean  to  the  mean  ;  like- 
wise the  other  extreme  shall  have  to  the  other  a  given 
ratio. 

Let  A,  B,  C  be  three  proportional  straight  lines,  and  D,  E,  F 
three  other  ;  and  let  the  ratios  of  A  to  D,  and  of  C  to  F  be 
given  ;  then  the  ratio  of  B  to  E  is  also  given. 

Because  the  ratio  of  A  to  D,  as  also  of  C  to  F  is  given,  the 

a  6r.  dat.  ratio  of  the  rectangle  A,   C  to  the  rectangle  D,   F  is  given  ^  ; 

b  17.  6.    but  the  square  of  B  is  equal  ^  to  the  rectangle  A,  C ;  and  the 

square  of  E  to  the  rectangle  ^^  D,  F  ;  therefore  the  ratio  of  the 

c58.dat. square  of  B  lo  the  square  of  E  is  given  j  wherefore  ^  also  the 

ratio  of  the  straight  line  B  to  E  is  given. 

Next,  let  the  ratio  of  A  to  D,  and  of  B  to  E  be  gi- 
ven ;  then  the  ratio  of  C  to  F  is  also  given. 

Because  the  ratio  of  B  to  E  is  given,   the  ratio  of      »     J.     X 

d  54;  dat.  the  square  of  B  to  the  square  of  E  is  given  ^  ;  there-     n    k    f 

fore  ^  the  ratio  of  the  rectangle  A,  C  to  the  rectangle 

D,  F  is  given  ;  and  the  ratio  of  the  side  A  to  the  side 

P  is  given ;  therefore  the  ratio  of  the  other  side  C  to 

eG5.  dat. the  other  F  is  given  e. 

Cou.  And  if  the  extremes  of  four  proportionals  have  to  the 
extremes  of  four  other  proportionals  given  ratios,  and  one  of 
the  means  a  given  ratio  to  one  of  the  means  ;  the  other  mean 
shall  have  u  given  ratio  to  the  other  mean,  as  may  be  shown  in 
the  same  manner  as  in  the  foregoing  proposition. 


DATA, 


42S 


PROP.  LXXII. 


82, 


IF  four  straight  lines  be  proportionals  ;  as  the  first  is 
to  the  straight  line  to  Avhich  the  second  has  a  given  ratio, 
so  is  the  third  to  a  straight  line  to  \\  hich  the  fourth  has  a 


given  ratio. 


Let  A,  B,  C,  D  be  four  proportional  straight  lines,  viz.  as 
A  to  B,  so  C  to  D  ;  as  A  is  to  the  straight  line  to  which  B  has 
a  given  ratio,  so  is  C  to  a  straight  line  to  which  D  has  a  given 
ratio. 

Let  E  be  the  straight  line  to  which  B  has  a  given 
ratio,  and  as  B  to  E,  so  make  D  to  F  ;  the  ratio  of 
B  to  E  is  given  ^ ,  and  therefore  the  ratio  of  D  to  F  ; 
and  because  as  A  to  B,  so  is  C  to  D  ;  and  as  B  to  E 
so  D  to  F ;  therefore,  ex  jcquali,  as  A  to  E,  so  is  A  li 
C  to  F ;  and  E  is  the  straight  line  to  which  B  has  a  C  U 
given  ratio,  and  F  that  to  v.'hich  D  has  a  given  ratio ; 
therefore  as  A  is  to  the  straight  line  to  which  B  has 
a  given  I'atio,  so  is  C  to  a  line  to  which  D  has  a  gi- 
ven ratio. 


a  Hyg. 


E 
F 


PROP.  LXXIIL 


8S. 


IF  four  straight  lines  be  proportionals ;   as  the  first  ^^^  ^• 
is  to  the  straight  line  to  which  the  second  has  a  given 
ratio,  so  is  a  straight  line  to  which  the  third  has  a  given 
ratio  to  the  foiirdi. 


Let  the   straight  line  A  be  to  B,   as  C  to  D ;   as  A  to  tlve 
straight  line  to  which  B  has  a  given  ratio,    so  is  a 
straight  line  to  which  C  has  a  given  ratio  to  D. 

Let  E  be  the  straight  line  to  which  B  has  a  given 
ratio,   and  as  B  to  E,  so  make  F  to  C;  because  the 
ratio  of  B  to  E  is  given,  the  ratio  of  C  to  F  is  gi-     ABE 
ven  :  and  because  A  is  to  B,  as  "^^  to  D  ;   and  as  B      F    C    D 
to  E,  so  F  to  C  ;  therefore,  ex  xquali  in  proportione 
perturbata  »,  A  is  to  E,  as  F  to  D  ;  that  is,  A  is  to  E  a  31  Si 

to  which  B  has  a  given  ratio,  as  F,  to  which  C  has 
a  given  ratio,  is  to  D. 

3  H 


426  EUCLID'S 


^-  PROP.  LXXIV. 


IF  a  triangle  has  a  given  obtuse  angle  ;  the  excess  of 
the  square  of  the  side  which  subtends  the  obtuse  angle, 
aboN'c  the  squares  of  the  sides  which  contain  it,  shall 
have  a  inven  ratio  to  the  triane-le. 

o  o 

Let  the  triangle  ABC  have  a  giren  obtuse  angle  ABC ;  and 
produce  the  straight  line  ('B,  and  from  the  point  A  draw  AD 
perpendicular  to  BC  :  the  excess  of  the  square  of  AC  abcv..  the 

a  12.  2.  squares  of  AB,  BC,  thatis-"*,  the  double  of  the  rectangle  con- 
tained by  DB,  BC,  has  a  given  ratio  to  the  triangle  ABC. 

Because  the  angle  ABC  is  given,  the  angle  ABD  is  also  gi- 
ven ;    and   the    angle   ADB   is   given  ;    wherefore   the   triangle 

b43.dat.  A13D  is  given  ^  in  species;    and  therefore  the  ratio   of  AD    to 

c  1.  6.  DB  is  given  :  and  as  AD  to  DB,  so  is  ^  the  rectangle  AD, 
BC  to  the  rectangle  DB,  BC  ;  wherefore  the  ratio  of  the  rect- 
angle AD,  BC  to  the  rectangle  DB,  BC  is  given,  as  also  the 
ratio  of  twice  the  rectangle  DB,  BC  to  the    A  E 

rectangle  AD,  BC:  but  the  ratio  of  the  rect- 
angle AD,  BC  to  the  triangle  ABC  is  gi- 

d  41. 1.    Yen,  because  it  is  double  ^  of  the  triangle  ; 
therefore  the  ratio  of  twice  the  rectangle 

e  9.  dat.   DB,    BC  to  the  triangle   ABC  is  given  «  ; 

and  twice  the  rectangle  DB,  BC  is  the  ex-     D     B  C 

cess  ^of  the  square  of  AC  above  the  squares  of  AB,  BC ;  there- 
fore this  excess  has  a  given  ratio  to  the  triangle  ABC. 

And  the  ratio  of  this  excess  to  the  triangle  AEC  may  be  found 
thus :  take  a  straiglit  line  EF  given  in  posititn  and  magnitude  ; 
and  because  the  angle  ABC  is  given,  at  the  point  F  of  the  straight 
line  EF,  make  the  angle  EFG  equal  to  the  angle  ABC  ;  pro- 
duce GF,  and  draw  EH  perpendicular  to  FG  ;  then  the  ratio  of 
the  excess  of  the  square  of  AC  above  the  squares  of  AB,  BC  to 
the  triangle  ABC,  is  the  same  with  the  ratio  of  quadruple  the 
straight  line  HF  to  HE. 

Because  the  angle  ABD  is  equal  to  the  angle  EFH,  and  the 
angle  ADB  to   LHF,    eacli  being  a  right   angle  ;    the   triangle 

£4.6.      ADB   is   equiangular    to    EHF  ;    therefore  *"  as  BD  to  DA,    so 

2Cor.4  5.  FH  to  HE  ;  and  as  quadruple  of  liD  to  DA,  so  is  s  quadru- 
ple of  FH  to  HE  :  but  as  twice  BD  is  to  DA,  so  is  =  twice 
the  rectangle  DB,    BC  to  the  rectangle  AD,    BC  ;    and  as  DA 

h  C»r.  5.  to  the  ha)f  of  it,  so  is  '^  the  rectunjile  AD,   BC  to  its  half  the 


DATA.  427 

triangle  ABC;  therefore,  ex  aquali,  as  twice  BD  is  to  the  half 
of  DA,  that  is,  as  quadruple  of  BD  is  to  DA,  that  is,  as  qua- 
di'plt^  of  FH  to  HE,  so  is  twice  the  rectangle  DB,  BC  to  the 
triangle  ABC. 

PROP.  LXXV.  65. 

IF  a  triangle  has  a  given  acute  angle,  the  space  l^y 
which  the  square  of  the  side  subtending  the  acute  angle, 
is  h.iis  than  the  squares  of  the  sides  which  contain  i^, 
sli;.'i  have  a  given  ratio  to  the  triangle. 

Let  the  triangle  ABC  have  a  given  acute  angle  ABC,  and  draw 
AD  perpendicular  to  BC,  the  space  by  which  the  square  of  AC 
is  i-ss  than  the  squares  of   ^B,   BC,  that  is  S  the  double  o^^the^  13.  2i 
rectangle  contained  by  CB,  BD,  has  a  given  ratio  to  the  triangle 
ABC. 

Because  tlse  angles  ABD,  ADB  are  each  of  them  given, 
the  tridngle  ABD  is  given  in  species;  and  therefore  the  ratio 
of  BD  to  DA  is  given  :  and  as  BD  to   DA,  A 

so  is  the  rectangle  CB,  BD  to  the  rectangle 
CB,  AD  ;  therefore  the  ratio  of  these  rect- 
angles is  given,  as  also  the  ratio  of  twice  the 
rectangle  CB,  BD  to  the  rectangle  CB,  AD  ; 
but  the  rectangle  CB,  AD  has  a  given  ratio 
to  its  half  the  triangle  ABC  ;  therefore  ^  the  B  D      C    1,9  j^^ 

ratio  of  twice  the  rectangle  CB,  BD  to  the  triangle  ABC  is  given  ; 
and  twice  the  rectangle  CB,  BD  is  ^  the  space  by  which  the 
square  of  AC  is  less  than  the  squartsof  VB,  BC  ;  therefore  the 
ratio  of  this  space  to  the  triangle  ABC  is  given  :  and  the  ratio 
may  be  found  as  in  the  preceding  proposition. 

LEMMA. 

IF  from  the  vertex  A  of  an  iscosceles  triangle  ABC,  any  straight 
line  AD  be  drawn  to  the  base  BC,  the  square  of  the  side  AB  is 
equal  to  the  rectangle  BD,  DC  of  the  segments  of  the  base  toge- 
ther with  the  square  of  AD  ;  but  if  AD  be  drawn  to  the  base 
produced,  the  square  of  AD  is  equal  to  the  rectangle  BD,  DC 
together  with  the  square  of  AB. 

Case  1.     Bisect  the  base  BC  in  E,  and  A 

join  AE  which  will  be  perpendicular  ^  to 
BC  ;  wherefore  the  square  of  AB  is  equal 
••to  the  squares  of  i\E,  EB  ;  but  the  square 
of  EB  is  equal  =^  to  the  rectangle  BD,  DC 
together  with  the  square  of  DE  ;  there- 
fore   the   square    of   AB    is    equal  to   the 


423 


EUCLID'S 


b  47. 1.  squares  of  AE,  ED,  that  is,  to  ^  the  square  of  AD,  together  with 
the  rectangle  BD,  DC  ;  the  other  case  is  shown  in  the  same  way 
by  6.  2.  Elem. 


67. 


PROP.  LXXVI. 


a5.&.32 
1. 

b  45.  dat 


c  50.  dat 
d  1.  6. 
e41    1. 
fr,7,  1. 
jr9,  dat. 


IF  a  triangle  have  a  ghen  angle,  the  excess  of  the 
square  of  the  straight  Une  which  is  equal  to  the  t\^^o  sides 
that  contain  the  gi^^en  angle,  above  the  square  of  the 
third  side,  shall  hdve  a  given  ratio  to  the  triangle. 

Let  the  triangle  ABC  have  the  given  angte  BAC,  the  excessof 
the  square  of  the  straightline  which  is  equal  to  BA,  AC  togeiher 
abovu  the  square  of  BC,  shall  have  a  given  ratio  to  the  triangle 
ABC. 

Pioduce  BA,  and  take  AD  equal  to  AC,  join  DC  and  pro- 
duce it  to  E,  and  through  the  point  B  draw  BE  parallel  to 
AC  ;  join  AE,  and  draw  AF  perpendicular  to  DC  ;  and  be- 
cause AD  is  equal  to  AC,  BD  is  equal  to  BE  :  and  BC  is 
drawn  from  the  vertex  B  of  the  isosceks  triangle  DBE  ;  there- 
fore, by  the  lemma,  the  square  of  BD,  that  is,  of  BA  and 
AC  together,  is  equal  to  the  rectangle  DC,  CE  together  with 
the  squai'e  of  BC  :  and,  therefore,  the  square  of  BA,  AC  to- 
gether, that  is,  of  BD,  is  greater  than  D 
the  square  of  BC  by  the  rectangk  DC, 
CE ;  and  this  rectangle  has  a  given 
ratio  to  the  ti'iangle  ABC :  because 
the  angle  BAC  is  given,  the  adjacent 
angle  CAD  is  given  ;  and  each  of  the 
angles  ADC,  DC  A  is  given,  for  each  -^ 
of  them  is  the  half  ^  of  the  given  angle  q  y^jL^ 
BAC  ;  therefore  the  triangle  aDC  is  "\1 
given  ^  in  species;  and  AI'  is  drawn  ^^K 
from  its  vertex  to  the  base  in    a   given 

angle  ;  wherefore  the  ratio  of  AF  to  the  base  CD  is  given  =  ;  and 
as  CD  to  At",  so  is'^  the  rectangle  DC,  CE  to  the  rectangle  AF, 
CE  ;  and  the  ratio  of  the  rectangle  AF,  CE  to  its  Iialf^  the  tri- 
angle  ACE  is  givtn  ;  therefor'.;  the  ratio  of  the  rectangle  DC, 
CE  to  the  triangle  ACE,  tliat  is*",  to  the  triangle  ABC,  is  given  sr 
and  the  rectangle  DC,  CE  is  the  excess  of  the  square  of  BA, 
AC  together  above  the  square  of  BC  :  therefore  the  ratio  of  this 
excess  to  the  triangle  ABC  is  given. 

Tne  ratio  wliich  the  rectangle  DC,  CE  has  to  the  triangle 
ABC  is  found  thusc  take  the   straight  line  GH  given  in  posi- 


DATA. 


42!9 


tion  and  magnitude,  and  at  the  point  G  in  GH  make  the  angle 
HGK  equal  to  the  given  angle  CAD,  and  take  GK  equal  to 
GH,  join  iiH,  and  draw  GL  perpendicular  to  it;  then  the  ratio 
of  HK  to  the  half  of  GL  is  the  same  with  the  ratio  of  the  rect- 
angle DC,  CE  to  the  triangle  ABC :  because  the  angles  HGK, 
DAC  at  the  vertices  of  th^  isosceles  triangles  GHK,  ADC  are 
equal  to  one  another,  these  triangles  are  sin\ilar  ;  and  because 
GL,  AF  are  perpendicular  to  the  bases  HK,  DC,  as  HK  to  GL, 
so  is*^  (DC  to  AF,  and  so  is)  the  rectangle  DC,  CE  to  the  rect- 
angle AF,  CE  ;  but  as  GL  to  its  half,  so  is  the  rectangle,  AF, 
CE  to  its  half,  which  is  the  triangle  ACE,  or  the  triangle  ABC  ; 
therefore,  ex  sequali,  HK  is  to  the  half  of  the  straight  line  GL, 
as  the  rectangle  DC,  CE  is  to  the  triangle  ABC. 

Cor.  And  if  a  triangle  have  a  given  angle,  the  space  by  whicli 
the  square  of  the  straight  line  which  is  the  difference  of  the 
sides  which  contain  the  given  angle  is  less  than  the  square  of 
the  third  side,  shall  have  a  given  ratio  to  the  triangle.  This  is 
demonstrated  the  same  way  as  tlie  preceding  proposition,  by 
help  of  the  second  case  of  the  lemma. 


C4.  6 


PROP.  LXXVII. 


L. 


IF  the  perpendicular  drawn  from  a  given  angle  of  a  SeeNote. 
triangle  to  the  opposite  side,  or  base,  has  a  gi^'en  ratio 
to  the  base,  the  triangle  is  given  in  species. 

Let  the  triangle  ABC  have  the  given  angle  BAC,  and  let  the 
perpendicular  AD  drawn  to  the  base  BC  have  a  given  ratio  to 
it,  the  triangle  ABC  is  given  in  species. 

If  ABC  be   an   isosceles  triangle,  it   is  evident  ^  that  if  anya5.&32'. 

1. 
G 


one  of  its  angles  be  given,  the  rest  are  also  given  ;  and  there- 
fore the  triangle  is  given  in  species,  without  the  consideration 
of  the  ratio  of  the  perpendicular  to  the  base,  whicli  in  this  case 
is  given  by  prop.  30. 

But  when  ABC   is  not  an  isosceles  triangle,  take  any  straight 
line  EF  given  in  position   and  magnitude;  and  upon  it  describe 


4S0 


EUCLID'S 


the  segment  of  a  circle  EGF  containing;  an  angle  equal  to  the 
given  angle  BAG,  draw  GH  bisecting  EF  at  right  angles,  and 
join  EG,  GF  :  then,  since  the  angle  EGF  is  equal  to  the  angle 
BAG,  and  that  EGF  is  an  isosceles  triangle,  and  ABC  is  not, 
the  angle  FEG  is  not  equal  to  the  angle  CBA  :  draw  EL  ma- 
king the  angle  FEL  equal  to  the  angle  CBA  ;  join  FL,  and 
draw  LM  perpendicular  to  EF  ;  then,  because  the  triangles  ELF, 
BAG  are  equiangular,  as  also  are  the  triangles  MLE,  DAB, 
as  ML  to  LE,  so  is  DA  to  AB  ;  and  as  LE  to  EF,  so  is  AB  to 
BC  ;  wherefore,  ex  squali,  as  LM  to  ILF,  so  is  AD  to  BC  ; 
and  because  the  ratio  of  AD  to  BC  is  given,  therefore  the  ratio 

b2.dat.  of  LM  to  EF  is  given  ;  and  EF  is  given,  wherefore  ^  LM  also 
is  given.  Complete  the  parallelogram  LMFK  ;  and  because  LM 
is  given,  FK  is  given  in  magnitude  ;  it  is  also  given  in  position, 

c30.  dat.  and  the  point  F  is  given,  and  consequently  "^  the  point  K  ;  and  be- 
cause through  K  the   straight  line  KL   is  drawn  parallel  to  EF 

d 51.dat.  which  is  given  in  position,  thereibreti  KL  is  given  in  position r 


B         Pv  D   C 


E         O        H       M       F 


and  the   circumference  ELF  is  given  in  position  ;  therefore  tha 
e  ^8.  dat.  point  L  is  given «".     And  because  the  points  L,  E,  F,  are  given, 
f  29.dat.tl^^  straight  lines  LE,  EF,  FL,  are  given  '  in  magnitude  ;  there- 
"42. dat. ^^^^   the    triangle  LEF   is  given   in   species  ?;  and  the  triangle 
ABC  is  similar  to  LEF,  wherefore  also  ABC  is  given  in  species. 
Because  LM  is  less  than  GH,  the  ratio  of  LM  to  EF,  that  is, 
the  given  ratio  of  AD  to  BC,  n\ust  be  less  than  the  ratio  of  GH 
to  EF,  which  the  straight  line,  in  a  segment  of  a  circle  contain- 
ing an  angle  equal  to  the  given  angle,  that  bisects  the   base  of 
the  segment  at  right  angles,  has  unto  the  base. 

CoR.  1.  If  two  triangles,  ABC,  LEF  have  one  angle  BAG 
equal  to  one  angle  ELF,  and  if  the  perpendicular  AD  be  to  the 
base  BC,  as  the  perpendicular  LM  to  the  b?,se  EF,  the  triangles 
ABC,  LEF  are  sinular. 

Describe  the  circle  EGF  about  the  triangle  ELF,  and  draw 
LN  pardlcl  to  EF,  join  EN,  NF,  and  draw  NO  perpeneicu- 
lar  to  EF  ;  because  the  angles  ENF,  ELF  are  equal,  and  that 


DATA.  431 

the  angle  EFN  is  equal  to  the  alternate  angle  FNL,  that  is,  t© 
the  angle  FEL  in  the  sume  segment  ;  therefore  the  triangle 
NEF  is  similar  to  LEF  ;  and  in  the  segment  EGF  there  can  be 
no  other  triangle  upon  the  base  EF,  which  has  the  ratio  of  its 
perpendicular  to  that  base  the  same  with  the  ratio  of  LM  or  NO 
to  EF,  because  the  perpendicular  must  be  greater  or  less  than 
LM  or  NO  ;  but,  as  has  been  shown  in  the  preceding  demon- 
stration, a  triangle  similar  to  ABC  can  be  described  in  the  seg- 
ment EGF  upon  the  base  EF,  and  the  ratio  of  its  perpendicular 
to  tiie  base  is  the  same,  as  was  there  shown,  with  the  ratio  of 
AD  to  BC,  that  is,  of  LM  to  EF ;  therefore  that  triangle  must 
be  either  LEF,  or  NEF,  which  therefore  are  similar  to  the  tri- 
angle ABC. 

Cor.  2.  If  a  triangle  ABC  has  a  given  angle  BAC,  and  if  the 
sitriiight  line  AR  drawn  from  the  given  angle  to  the  opposite 
sidt  EC,  ill  a  given  angle  ARC,  has  a  given  ratio  to  BC,  the  tri- 
angle ABC  is  given  ir»  species. 

Draw  AD  perpendicular  to  BC  ;   therefore  the  triangle  ARD 
is  gi\en  in  species  ;   wherefore  the  ratio  of  AD  to  AR  is  given  : 
and  the  ratio  of  AR  to  BC  is  given,  and  consequently  ^  the  ratiohS-dat. 
of  AD  to  BC  is  given  ;  and  the  triangle  ABC  is  therefore  given 
in  specic^s'.  177.  dat. 

CoR.  3.  If  two  triangles  ABC,  LEF  have  one  angle  BAC 
equal  to  one  angle  ELF,  and  if  straight  lines  drawn  from  these 
angles  to  the  bases,  making  with  them  given  and  equal  angles, 
have  the  same  ratio  to  the  bases,  each  to  each  ;  then  the  trian- 
gles are  similar  \  for  having  drawn  perpendiculars  to  the  bases 
from  the  equal  angles,  as  one  perpendicular  is  to  its  base,  so  is  .  ^ 
the  other  to  its  buse  i^  ;  wherefore,  by  Cor.  1,  the  triangles  are  J^  ^ 32  5. 
similar. 

A  triangle  similar  to  ABC  may  be  found  thus  :  having  de- 
scribed the  segment  EGF,  and  drawn  the  straight  line  GH,  as 
was  oircctcd  in  the  proposition,  find  IK,  which  has  to  EF  the 
given  ratio  of  AD  to  BC  ;  and  place  FK  at  right  angles  to  EF 
from  ihc  point  F  ;  then  because,  as  has  been  shown,  the  ratio 
of  AD  to  BC,  that  is  of  FK  to  El',  must  be  less  than  the  ratio 
of  GH  to  EF  ;  therefore  FK  is  less  than  GH;  and  consequently 
the  parallel  to  EF,  drawn  through  the  point  K,  must  meet  the 
circumference  of  the  segment  in  two  points  :  let  L  be  either  of 
them,  and  join  EL,  LF,  and  draw  LM  perpendicular  to  EF  : 
then,  because  the  angle  BAC  is  eqiial  to  the  angle  ELF,  and 
that  aD  is  to  Be,  .as  KF,  that  is  LM,  to  EF,  the  triangle  ABC 
is  similar  to  the  triangle  LEF,  by  Cor.  l. 


439 


EUCLID'S 


80. 


PROP.  LXXVIII. 


IF  a  triangle  ha^e  one  angle  given,  and  if  the  ratio 
of  the  rectangle  of  the  sides  which  contain  the  given  an- 
gle to  the  square  of  the  third  side  be  given,  the  triangle 
is  given  in  species. 


Let  the  triangle  ABC  have  the  given  angle  BAC,  and  let 
the  ratio  of  the  rectangle  BA,  AC  to  the  square  of  BC  be  gi* 
ven  ;  the  triangle  ABC  is  given  in  species. 

From  the  point  A,  draw  AD  perpendicular  to  BC,  the  rect- 
a  41.  1.    angle  AD,  BC  has  a  given  ratio  to  its  half^  the  triangle  ABC; 
and  because  the  angle  BAC  is  given,  the  I'atio   of  the  triangle 
bCor62.  ABC  to   the  rectangle  BA,   AC   is  given  "^  ;  and   by   the   hypo- 
dat.  thesis,  the  ratio  of  the  rectangle  BA,  AC  to  the  square  of  BC  is 

c9.dat.  given;  therefore  "=  the  ratio  of  the  rectangle  AL>,  BC  lo  the 
dl.  6.  square  of  BC,  that  is  d,  thei  ratio  of  the  straight  line  AD  to  BC 
e77.dat.  is  given  ;  wherefore  the  triangle  ABC  is  given  in  species  e. 

A    triangle    similar   to    ABC    may    be    found  thus  :    take    a 
straight   line    EF   given    in  position  and   magnitude,  and  make 
the  angle  FEG  equal  to  the  given  angle  BAC,  and  draw  FH 
perpendicular  to  EG,  and  BK  perpendicular  to  AC ;  therefore 
the  triangles  ABK,  EFH 
are  similar,  and  the  rect- 
angle   AD,    BC,    or    the 
rectangle  BK,  AC  which 
is  equal  to  it,    is    to  the 
rectangle  BA,  AC,  as  the 
straight   line  BK   to  BA, 
that  is,  as  FH  to  FE.  Let 
the  given  ratio  of  the  rect- 
angle BA,  AC  to  the  square  of  BC  be  the  same  with  the  ratio  of 
the  straight  line  EF  to  FL  ;  therefore,  ex  xquali,  the  ratio  of  the 
rectangle  AD,  BC  to  the  square  of  BC,  that  is,  the  ratio  of  the 
straight  line  AD  to  BC,  is  the  same  uith  the  ratio  of  HF  to  FL ; 
and  because  AD  is  not  greater  than  the  straight  line  MN  in  the 
segment  of  the  circle  described  about  the  triangle   ABC,  which 
liisects  BC  at  right  angles  ;  the  ratio  of  AD  to  BC,  that  is,  of 
HF  to  FL,  must  not  be  greater  than  the  ratio  of  MN  to  BC  : 
let  it    be    so,    and,    by    the    77th    dat.    find    a    triangle    OPQ 
which   has  one    of   its    angles    POQ    equal    to    the   given    an- 
gle BAC,  and  the  ratio  of  the    pL-rpendicular  OR,  drawn  from 
that  angle  to  the  base  PQ  the  same  with  the  ratio  of  HF  to  FL  ; 
then  the  triangle  ABC  is  similar  to  OPQ:  because,  as  has  bee» 


B  D 


J> 


c  10.  dat. 
d  7.  dat. 


DATA.  433 

shown,  the  ratio  of  AD  to  BC  is  the  same  with  the  ratio  of 
(HF  to  FL,  that  is,  by  the  construction,  with  the  ratio  of)  OR 
to  PQ  :  and  the  angle  BAC  is  equal  to  tlie  angle  POQ  ;  there- 
fore the  triangle  ABC  is  similar  f  to  the  triangle  POQ.  f  1-  Cnr. 

77.  dat. 

Ot/termise, 

Let  the  triangle  ABC  have  the  given  angle  BA.C,  and  let 
the  ratio  of  the  rectangle  BA,  AC  to  the  square  of  BC  be  given  ; 
the  triangle  \^C  is  given  in  species. 

Because  the  angle  BAC  is  given,  the  excess^  of  the  square 
of  both  the  sides  BA,  AC  together  above  the  square  of  the 
third  side  BC  has  a  given  ^  ratio  to  the  triangle  ABC.  Let  the  a  76,  dat. 
figure  D  be  equal  to  this  excess  ;  thevefoie  the  ratio  of  D  to 
the  triangle  ABC  is  given  :  and  the  ratio  of  the  triangle  ABC 
to  the  rectangle  BA,  AC   is   given'',  because  BAC  is  a  given  I)  Cor.  62. 

angle  ;  and  the  rectangle  BA,  AC  has  ^  ^ dat 

a    given    ratio    to    the  square  of  BC  : 

wherefore  ^  the  ratio  of  D  to  the  square 

of  BC  is  given  ;  and,  by  composition  '^ 

the  ratio  of  the  space  D  together  with 

the  square  of  BC  to  the  square  of  BC     B  C 

is  given  ;  but  D  together  with  the  square  of  BC  is  equal  to  the 

square  of  both  BA  and  AC  together;  therefore  the  ratio  of  the 

square  of  BA,  AC  together  to  the  square  of  BC  is  given  ;  and  the 

ratio  of  BA,  AC  together  to   BC  is  therefore  given  e;  and  thee59.dat 

angle  BAC  is   given,  wherefoi'e  *"  the  triangle  ABC  is  given  inf48.dat. 

species. 

The  composition  of  this,  which  depends  upon  those  of  the 
76th  and  48th  propositions,  is  more  complex  than  the  preced- 
ing composition,  which  depends  upon  that  of  prop.  77,  which 
is  easy. 


PROP,  LXXIX.  K. 

IF  a  triangle  have  a  given  angle,  and  if  the  straight  SeeNotc. 
line  drawn  from  that  angle  to  the  base,  making  a  given 
angle   with   it,    divides  the   base  into  segments   m  hich 
have  a  given  ratio  to  one  another  ;  the  triangle  is  given 
in  species. 

Let  the  triangle  ABC  have  the  given  angle  BAC,  and  let 
the  straight  line  AD  drawn  to  the  base  BC  making  the  given 
angle  ADB,  divide  BC  into  the  segments  BD,  DC  which  have 

31 


434  EUCLID'S 

a    given    ratio    to  one  anotl:ier ;    the  triangle  ABC  is  given  In 
species. 

a  5.  4.  Describe  ^  the  circle  BAC  about  the  triangle,  and  from  its 
cc;Ure  E,  draw  EA,  EB,  EC,  ED  ;  becuuse  the  angle  BAC  is 

b  20.  3.  given,  the  angle  BEC  at  the  centre,  which  is  the  double  ^  of  it, 
is  given.     And   the  ratio  of  BE   to    EC   is  given,  because   they 

c  44.  dat.  ^[-(j  equal  to  one  another;  therefore'^  the  triangle  BEC  is 
given  in  species,  and  the  ratio   of  EB  to  BC  is  given  ;  also  the 

d7.dat.  i-^tio   of  CB  to  BD  is  given '^  because  the  ratio  of  BD  to   DC 

e9.dat.  ig  oiven  ;  therefore  the  ratio  of  EB  to  BD  is  given  <=,  and  the 
angle  EBC  is  given,  wherefore  the  triangle  EBD  is  given  < 
in  species,  and  the  ratio  of  LB,  that  is,  of  EA  to  ED,  is  there- 
fore given  ;  and  the  angle  EDA  is  given,  because  each  of  the 
angles  BDE,  BDA   is    given;  therefore   the    triangle    AED  is 

f  4?".  dat.given'*"  in  species,  and  the  angle  AED  gi- 
ven ;  also  the  angle  DEC  is  given,  because 
each  of  the  angles  BED,  BEC  is  given  ; 
therefore  the  angle  AEC  is  given,  and  the 
ratio  of  EA  to  EC,  which  are  equal,  is  gi- 
ven ;  and  the  triangle  AEC  is  therefore 
given  ^'  in  species,  and  the  angle  ECA  gi- 
ven ;  and  the  angle  ECB  is  given,  wherefore  the  angle  ACB  is 

543.  dat.  given,  and  the  angle  BAC  is  also  given  ;  therefore  s  the  triangle 
ABC  is  given  in  species. 

A  triangle  similar  to  ABC  may  be  found,  by  taking  a  straight 
line  given  in  position  and  magnitude,  and  dividing  it  in  th^ 
given  ratio  which  the  segments  BD,  DC  are  required  to  have 
to  one  another;  then,  if  upon  that  straight  line  a  segment  of  a 
circle  be  described  containing  an  angle  equal  to  the  given,  angle 
BAC,  and  a  straight  line  be  drawn  from  the  point  of  division  in 
un  angle  equal  to  the  given  angle  ADB,  and  from  the  point 
wliere  it  meets  the  circumference,  straight  lines  be  drawn  to 
the  extremity  of  the  first  line,  these,  together  with  the  first  line, 
shall  contain  a  triangle  similar  to  ABC,  as  may  easily  be  shown. 
The  demonstration  may  be  also  made  in  the  manner  of  that 
of  the  77th  prop,  and  that  of  the  77th  may  be  made  in  the  man- 
ner of  this. 


I     A 

a  26.  1, 

/N 

:^ 

•  43.  dat 

B         D 

C 

bU.  cUt, 

DATA.  433 


PROP.  LXXX. 


IF  the  sides  about  an  angle  of  a  triangle  have  a  given 
ratio  to  one  another,  and  if  the  perpendicuiar  drawn  iiom 
that  angle  to  the  base  has  a  given  ratio  to  the  I'^asc ;  tlie 
triangle  is  given  in  species. 

Let  the  sides  BA,  AC,  about  the  angle  BAG  of  the  triangk- 
ABC  have  a  given  ratio  to  one  another,  and  let  the  perpendicu- 
lar AD  have  a  given  ratio  to  the  base  BC  ;  the  triangle  ABC  is 
given  in  species. 

First,  let  the  sides  AB,  AC  be  equal  to  one  another,  there- 
fore the  perpendicular  AD  bisecls  -^  the  base 
BC  ;  and  the  ratio  of  AD  to  BC,  and  there- 
fore to  its  half  DB,  is  given  ;  and  the  angle 
ADB  is  given  ;  wherefore  the  triangle  *  ABD, 
and  consequently  the  triangle  ABC,  is  given '^ 
in  species. 

But  let  the  sides  be  unequal,  and  BA  be  greater  than  AC  ; 
and  make  the  angle  Ci\  E  equal  to  the  angle  ABC;  becaiuie 
the  angle  AEB  is  common  to  the  triangles  AEB,  CEA,  they 
.-.re  similar;  therefore  as  AB  to  BE,  so  is  CA  lo  AE,  imd.  by 
permutation,  as  BA  to  AC,  so  i»  BE  to  EA,  and'bo  is  EA 
to  EC  ;  and  the  ratio  of  BA  to  AC  is  given,  therefore  the 
ratio  of  BE  to  EA,  and  tiie  ratio  of  EA  to  EC,  as  also  the 
ratio  of  BE  to  EC  is  given  = ;  wherefore  the  rat-.o  of  EB  to  c  9.  dat. 
BC  is  given '^ ;  and  the  mtio  of  AD  lo  BC  Ad6.dat. 

is  given  by  the  hypothesis,  therefore  '^  the 
ratio  of  AD  to  BE  is  given  ;  and  the  ratio 
BE  to  EA  was  shown  to  be  given  ;  where- 
fore the  ratio  of  AD  to  A.]L  is  given,  and 
ADE  is  a  right  angle,  therefore  the  triangle 
ADE  is  given  e  in  species,  and  the  angle  AEB  given  ;  thera-e46.dat. 
tio  of  BE  to  EA  is  likewise  given,  therefore  ^  the  lii.Uigle  ABE 
is  given  in  species,  and  consequently  the  angle  E -Mi,  as  also 
the  angle  ABE,  that  is,  the  angle  CAE,  is  given;  iheriibre  the 
angle  BAC  is  given,  and  the  angle  ABC  beirig  also  given,  the 
triangle  ABC  is  given  ^  in  species.  f  43.  dar. 

How  to  find  a  triarigle  v.hich  shall  have  the  things  which 
are  mentioned  to  be  given  in  the  proposition,  is  evident  in 
the  first  case ;  and  to  find  it  tlie  more  easily  in  tlie  other 
case,  it  is   to  be  observed  that,  if  the  strai^lu  line  EF  equal   to 


436 


lUCLID'S 


6.2. 


EA  be  placed  in  EB  towards  B,  the  point  F  divides  the  base 
BC  into  the  segments  BF,  FC  which  have  to  one  another  the 
ratio  of  the  sides  BA,'  AC,  because  BE,  EA  or  EF,  and 
19-  5.  EC  were  shown  to  be  proportionals,  therefore  *  BF  is  to  FC 
as  BE  to  EF,  or  EA,  that  is,  as  BA  to  AC  ;  and  AE  cannot 
be  less  than  the  altitude  of  the  triangle  ABC,  but  it  may  be 
equal  to  it,  which,  if  it  be,  the  triangle,  in  this  case,  as  also 
the  ratio  of  the  sides,  may  be  thus  found  ;  having  given  the 
ratio  of  the  perpendicular  to  the  base,  take  the  straight  line 
GH  given  in  position  and  magnitude,  for  the  base  of  the  tri- 
angle to  be  found  ;  and  let  the  given  ratio  of  the  perpendicu- 
lar to  the  base  be  that  of  the  straight  line  K  to  GH,  that  is,  let 
K  be  equal  to  the  perpendicular ;  and  suppose  Gi^H  to  be  the 
triangle  which  is  to  be  found,  therefore  having  made  the  angle 
HLM  equal  to  LGH,  it  is  required  that  LM  be  perpendicular  to 
GM,  and  equal  to  K  ;  and  because  GM,  ML,  MH  are  propor- 
tionals, as  was  shown  of  BE,  EA,  EC,  the  rectangle  GMH  is 
equal  to  the  square  of  ML.  Add  the  common  square  of  NH, 
(having  bisected  GH  in  N),  and  the  square  of  SM  is  equal  e  to 
the  squares  of  the  given  straight  lines  NH  and  ML  or  K  ;  there- 
fore the  square  of  NM  and  its  side  NM,  is  given,  as  also  the 
point  M,  viz.  by  taking  the  straight  line  NM,  the  square ■  of 
■which  is  equal  to  the  squares  of  NH,  ML.  Draw  ML  equal  to 
K,  at  right  angles  to  GM  ;  and  because  ML  is  given  in  position 
and  magnitude,  therefore  the  point  L  is  given,  join  LG,  LH ; 
then  the  triangle  LGH  is  that  which  was  to  be  found,  for  the 
sq':'.ire  of  NM  is  equal  to  the  squares  of  NH  and  ML,  and  taking 
awav  the  common   square  of 

^  L    K 

S 

3 


NH,  the  rectangle  GMH  is 
equal  s  to  the  square  of 
ML  :  therefore  as  GM  to 
ML,  so  is  ML  to  MH,  and 
h  6.  6.  ^'^^  triangle  LGM  is  ^^  there- 
fore equicingular  to  HLM, 
and  the  angle  HLM  equal  to  G  NQ  H  M  P 
the  angle  LGM,  and  the  straight  line  LM,  drawn  from  the 
vertex  of  the  triangle  making  the  angle  HLM  equal  to  LGK,  is 
perpendicular  to  the  base  and  equal  to  the  given  straight  line  K, 
as  was  required  ;  and  tiie  ratio  of  the  sides  GL,  LH  is  the  same 
with  the  ratio  of  CM  to  ML,  that  is,  with  the  ratio  of  the 
straight  line  which  is  made  up  of  GN  the  half  of  the  given  base 
and  of  NM,  the  square  of  which  is  equal  to  the  squares  of  GN 
and  K,  to  the  straight  line  K. 

And  whether  this  ratio  of  GM  to  ML  is  greater  or  less 
than  the  ratio  of  the  sides  of  any  other  triangle  upon  the  base 
GH,  and  of  which   the   altitude  is  equal  to  the  straight  line  K, 


DATA.  '  437 

fcliat  is,  the  vertex  of  which  is  in  the  parallel  to  GH  drawn 
through  the  point  L,  may  be  thus  found.  Let  OGH  be  any 
such  triangle,  and  draw  OP,  making  the  angle  HOP  equal  to 
the  angle  OGH  ;  therefore,  as  before,  GP,  PO,  PH  are  pro- 
portionals, and  PO  cannot  be  equal  to  LM,  because  the  rect- 
angle GPH  would  be  equal  to  the  rectangle  GMH,  which 
is  impossible  ;  for  the  point  P  cannot  fall  upon  M,  because  O 
would  then  fall  on  L  ;  nor  can  PO  be  less  than  LM,  therefore 
it  is  greater  ;  and  consequently  the  rectangle  GPH  is  greater 
than  the  rectangle  GMH,  and  the  straight  line  GP  greater 
than  GM:  therefore  the  ratio  of  GM  to  MH  is  greater  than 
the  ratio  of  GP  to  PH,  and  the  ratio  of  the  square  of  GM  to 
the  square  of  ML  is  therefore'  greater  than  the  ratio  of  the  >  2  Cor. 
square  of  GP  to  the  square  of  PO,  and  the  ratio  of  the  straight  2^*  *• 
line  GM  to  ML,  greater  than  the  ratio  of  GP  to  PO.  But  as 
GM  to  ML,  so  is  GL  to  LH  ;  and  as  GP  to  PO,-so  is  GO  to 
OH  ;  therefore  the  ratio  of  GL  to  LH  is  greater  than  the  ratio 
of  GO  to  OH  ;  wherefore  the  ratio  of  GL-to  LH  is  the  greatest 
of  all  others  ;  and  consequently  the  given  ratio  of  the  greater 
side  to  the  less,  must  not  be  greater  than  this  ratio. 

But  if  the  ratio  of  the  sides  be  not  the  same  with  this  great- 
est ratio  of  GM  to  ML,  it  must  necessarily  be  less  than  it : 
let  any  less  ratio  be  given,  and  the  same  things  being  suppo- 
sed, viz.  that  GH  is  the  base,  and  K  equal  to  the  altitude  of 
the  triangle,  it  may  be  found  as  follows.  Divide  GH  in  the 
point  Q,  so  .that  the  ratio  of  GQ  to  QH  may  be  the  same 
with  the  given  ratio  of  the  sides  ;  and  as  GQ  to  QH,  so  make 
GP  to  PQ,  and  so  will  f  PQ  be  to  PH  ;  wherefore  the  square  f  19.5.;; 
of  GP  is  to  the  square  of  PQ,  as'  the  straight  line  GP  to 
PH  :  and  because  GM,  ML,  MH  are  proportionals,  the  square 
of  GM  is  to  the  square  of  ML,  as'  the  straight  line  GM  to  MH  : 
but  the  ratio  of  GQ  to  QH,  that  is,  the  ratio  GP  to  PQ, 
is  less  than  the  ratio  of  GM  to  ML  ;  and  therefore  the  ratio 
of  the  square  of  GP  to  the  square  of  PQ  is  less  than  the  ratio 
of  the  square  of  GM  to  that  of  ML  ;  and  consequently  the 
ratio  of  the  straight  line  GP  to  PH  is  less  than  the  ratio  of 
GM  to  MH  ;  and,  by  division,  the  ratio  of  GH  to  HP  is  less 
than  that  of  GH  to  HM ;  wherefore  k  the  straight  line  HP  is  k  10. 5. 
greater  than  HM,  and  the  rectangle  GPH,  that  is,  the  square 
of  PQ,  greater  than  the  rectangle  GiMH,  that  is,  than  the 
square  of  ML,  and  the  straight  line  PQ  is  therefore  greater 
than  ML.  Draw  LR  parallel  to  GP  :  and  from  P  draw  PR  at 
right  angles  to  GP  :  because  PQ  is  greater  than  ML  or  PR, 
the  circle  described  from  the  centre  P,  at  the  distance  PQ, 
must  ntcessarily  cut  LR  in  two  points  j  let  these  be  O,  S,  and 


43S  EUCLID'S 

join  OG,  OK;  SG,  SIT:  each  of  the  triangles  OGH,  SGH 
have  the  things  mehi'ioned  lo  be  given  in  the  proposition  : 
join  OP,  SP;  and  beciiuse  as  GV  %  P'Q,  or  PO,  so  is  PO 
to  PH,  the  triangle  OGP  is  ecpiiingular  to  HOP  ;  as,  there- 
fore OG  lo  GP,  so  is  no  to  OP;  and,  bv  permutation,  as 
bo  to  OH  so  is  GP  to  PO,  or  'PQ  :  and  Jo  is  (iQ  lo  QH  : 
therefore  the  triangle  OCiH  has  the  ratio  of  its  sicits  GO,  OH 
tlVe  same  with  the  given  ratio  of  GQ  to  QH  :  and  the  perpendi- 
cular has  lo  the  base  the  given  ratio  .of  K  to  GH,  because  the 
perpendicular  is  equal  to  LMor  K  :  the  like  may  be  shown  in 
tiie,  same  way  of  the  triangle  SGPI. 

This  construction  by  which  the  triangle  OGH  is  found,  is 
shorter  than  that  which  would  be  deduced  from  the  demonstra- 
tion of  the  datum,  by  reason  that  the  base  GH  is  given  in 
position  and  magnitude,  v/hich  w.as  not,  supposed  in  the  de- 
monstraticn  :  the  same  thing  is  to  be  observed  in  the  next  pro- 
position. 


M.  PROP.  LXXXI. 


IF  the  sides  about  an  angle  of  a  triangle  be  unequal 
and  ha^'C  a  given  ratio  to  one  another,  and  if  the  perpen- 
dicular from  tiiat  angle  to  ,tbe  base,  di\ides  it  into  seg- 
ments that  ha've  a  gi^en  i-atio  to  one  another,  the  triaii- 
gle  is  given  in  species. 


Let  ABC  be  a  triangle,  th.e  sides  of  which  about  the  angle 
BAG  are  unequal,  and  have  a  given  ratio  to  one  another,  and 
let  the  perpendicular  AD  to  the,  base  BC  -livide  it  into  the  seg- 
ments BD,  DC,  which  have  a  given  ratio  to  one  another,  the 
triangle  ABC  is  given    in  species. 

Let  AB  be  greater   than    AC,  and    make    the   angle    CAE 

equal  to  the   angle    ABC  ;  imd  because  the  angle  ALB  is  com- 

a4  6'      mon   to  the   trianglis   ABE,  CAE,   they   are  =^  equiangular  to 

one  another  :  thercfcre  as   AB   lo    BE,  so   is  CA  to  AE,  and, 


DATA, 


4sa 


by  permutaiiOi),  as  AB  lo  AC,  so  BE  lo 
EA,  aiul  so  is  EA  to  EC  :  but  the  rutio  of 
BA.  lo  7\C  is  ;4iven,  therefore  the  ratio 
of  BE  to  EA,  as  also  the  ratio  of  EA  to 
EC  is  given  ;  wherefore''  the  ratio,  of 
BE  to  i.C,  as  also*^  the  ratio  of  EC  to 
CB  is  givtn  :  and  the  ratio  of  BC  to  CD: 
is  given'!,  because  the  ratio  of  ^  BD  to 
DC  is  given  ;  therefore  ^  the  ratio  of  EC 
to  CD  is  given,  and  consequenlly ''.  tlic 
ratio  of  DE  to  EC  :  and  the  ratio  of  EC  Q.  K,  L  II      N 

to  EA  was  shown  to  be  given,  therefore  ^  the.  ratio, of  DE  to.  EA 
is  given  ;  and  ADE  is  a  right  angle,  wherefore  «  the  triajigle 
ADE  is  given  in  species,  and  the  angle  AED  given  :  aod  the 
ratio  of  CE  to  EA  is  given,  therefore  f  the  triungle  AEG  is  gj^ 
ven  in  species,  and  consequently  the  angle  ACE  is  given,  as 
also  the  adjacent  angle  ACB.  In  the  same  manner,  because  the 
ratio  of  BE  to  EA  is  given,  the  triangle  BEA  is  given  in  spe- 
cies, and  the  angle  ABE  is  therefore  given:  a^^d  the  angle 
ACB  is  given  ;  wherefore  the  triangle  ABC  is  given  s  in  spe- 
cies. 

But  the  ratio  of  the  greater  side  BA  to  the  other  AC  must 
be  less  than  the  ratio  of  the  greater  segment  BD  to  DC  :  be- 
cause the  square  of  BA  is  lo  the  square  of  AC,  as  the  squares 
of  BD,  DA  to  the  squares  of  DC,  DA;  and  the  squares  of 
BD,  DA  have  lo  the  squares  of  DC,  DA  a  less  ratio  than  the 
square  of  BD  has  to  the  square  of  DC  t)  because  the  square  of 
BD  is  greater  than  the  square  of  DC  ;  therefore  tlie  square  of 
BA  has  lo  the  squ;\re  pf  AC  a  less  ratio  than  the  square  of  BD 
has  lo  that  of  DC  :  and  consequently  the  ratio  of  BA  to  AC  ;?i 
less  than  the  ratio  of  BD  to  DC. 

This  being  premised,  a  triangle  which  shall  have  the  thing-; 
mentioned  to  be  given  in  the  proposition,  and  to  which  the 
triangle  ABC  is  similar,  may  be  found  thus:  take  a  straight 
line  GH  given  in  position  and  magnitude,  and  divide  it  in  K, 
so  that  the  ratio  of  CK  to  KM  may  be  the  same  with  the  givcrt 
ratio  of  BA  to  AC;  divide  also  GH  irv  L,  so  that  the  ratio 
of  GL  to  Lli  may  be  the  same  with   the  given  ratio  of  BD    to 


e  4&;  da,t, 
f  44.  dat. 


sr  43.  dar. 


t  If  A  be  greater  then  B,  and  C  any  third  magnitude;  then  A  and  C  toge- 
ther have  to  B  and  C  together  a  less  ratio  than  A  has  to  B. 

Let  A  be  to  B  as  C  to  D,  and  because  A  is  greater  than  B,  C  is  greater 
than  D  :  but  as  A  is  to  B,  so  A  and  C  to  B  and  D ;  and  A  and  C  have  to  B 
and  C  a  less  fzfio  than  A  and  C  have  to  B  and  D,  because  C  is  greater  thui>. 
P,  thejefore  A  ai-d  C  have  to  B  and  C  a  less  ra-io  'ban  A  to  3. 


440  EUCLID'S 

DC,  and  draw  LM  at  right  angles  to  GH  :  and  because  the 
ratio  of  the  sides  of  a  triangle  is  less  than  the  ratio  of  the  seg- 
ments of  the  base,  as  has  been  shown,  the  ratio  GK  to  KH 
is  less  than  the  ratio  of  GL  to  LH :  wherefore  the  point  L 
must  fall  betwixt  K  and  H :  also  make  as  GK  to  KH,  so  GN 
h  19.  5  .to  NK,  and  so  shall  ^  NK  be  to  NH.  And  from  the  centre  N, 
at  the  distance  NK,  describe  a  circle,  and  let  its  circumference 
meet  LM  in  O,  and  join  OG,  OH  ;  then  OGH  is  the  triangle- 
v/hich  was  to  be  described  :  because  GN  is  to  NK,  or  NO,  as 
NO  to  NH,  the  triangle  OGN  is  equiangular  to  HON;  there- 
fore as  OG  to  GN,  so  is  HO  to  ON,  and,  by  permutation,  as 
GO  to  OH,  so  is  GN  to  NO,  or  NK,  that  is,  as  GK  to  KH, 
that  is,  in  the  given  ratio  of  the  sides,  and  by  the  construction, 
GL,  LH  have  to  one  another  the  given  ratio  of  the  segments  of 
the  base. 


60.  PROP.  LXXXII. 


IF  a  parallelogram  given  in  species  and  magnitude 
be  increased  or  diminished  by  a  gnomon  given  in 
magnitude,  the  sides  of  the  gnomon  are  given  in  mag- 
nitude. 

First,  Let  the  parallelogram  AB  given  in  species  and  magni- 
tude be  increased  by  the  given  gnomon  ECBDFG,  each'of  the 
straight  lines  CE,  DP  is  given. 

Because  AB  is  given  in  species  and  magnitude,  and  that  the 
gnomon  ECBDFG  is  given,  therefore  the  whole  space  AG 
is  given  in  magnitude  :  but  AG  is  also  given  in  spieces,  be- 
"S.dcf.  cause  it  is  similar^  to  AB :  therefore  the  sides  of  AG  are  gi- 
|2.andven'' ;  each  of  the  straight  lines  AE,  AF 
24.  o.  jg  therefore  given  ;  and  each  of  the  straight 
b  60.  dat.  lines  CA,  AD  is  given  S  therefore  each  of 
c  4.  dat.  the  remainders  EC,  DF  is  given  <^, 

Next,  let  the  parallelogram  AG  given  in 
species  and  magnitude,  be  diminished  by 
the  given  gnomon  ECBDFG,  each  of  the 
straight  lines  CE,  DF  is  given. 

Because  the   parallelogram  AG  is  given. 


H 


as  also  its  gnomon  ECBDFG,  the  remaining  space  AB  is  given 


DATA. 


AU 


in  magnitude  :  but  it  is  also  given  in  species  :  because  it  is  simi- 
lar ^  to  AG  ;  thereibrt:  ^  its  sides  CA,  AD  are  given,  and  each  oi^ 
the  straJLjht  lines  EA,  AF  is  given  ;    therefore  EC,  DF  are  each 
of  them  '^i^en. 

The  gnomon  and  its  sides  CE,  DF  may  be  found  thus  in  the 
firsi  case.  Let  H  be  the  given  sp.ict  to  which  the  gnomon  must 
be  made  equid.  and  find  ^  a  parallelogram  simil  a-  to  AB  and  d 
equal  to  the  figures  AB  and  H  togetlier,  and  place  its  sides  AE, 
AF  from  the  point  A,  upon  the  straight  lines  AC,  AD,  and  com- 
plete the  parallelogram  AG  which  is  about  the  same  diunieter«'e 
with  AB  ;  because  therefore  AG  is  equal  to  both  AB  and  H,  take 
away  the  common  p.\rt  AB,  the  remaining  gjiomon  ECBDFCr 
is  equal  to  the  remaining  figure  H  ;  therefore  a  gnomon  equal 
to  H,  and  iis  sides  CE,  DF  are  found  :  and  in  like  manner  they 
may  be  found  in  the  other  case,  in  wiiich  the  given  figure  H  must 
be  less  than  the  fi:^ure  FE  from  which  it  is  to  be  taken. 


25.0. 


26.  6. 


PROP.  LXXXIII. 


58. 


IF  a  parallelogram  equal  to  a  given  space  be  applied 
to  a  given  straight  line,  deficient  by  a  parallelogram 
given  in  species,  the  sides  of  the  defect  are  gi\en. 


Let  the  parallelogram  AC  equal  to  a  given  space  be  applied 
to  the  given  straight  line  AB,  deficient  by  the  parallelogram 
BDCL  given  in  species',  each  of  the  straight  lines  CD,  DB  are 
given. 

Bisect  AB  in  E  ;    therefore  EB  is  given  in   magnitude  :   upon 
EB  describe  ^  the  parallelogram  EF  similar  to  DL  and  similarly  a  18.  Cr. 
placed  ;  therefore  EF  is  given  in    species,  and  G     H  F 

is  about  the    same  diameter  ^  with  DL;  let  f;:^^     |      ]     b  2fi.  6. 

BCG    be   the  diameter,    and    construct  the 
figure  ;  therefore,  because  the  figure  EF  gi- 
ven in  sj  erii  s   is  described    upon  the   cTIVl. 
straight  line  EB,  EF  is  given  ^  in  magnitude,  ^^ 
and  tiie  gnomon  ELti  is   equal t^  to   me   gi- 
ven figure    I'    :  tUerclbrc  ^  since  EF  is  diminished  by  the  givei 
gnomoi'  ELH,  the   sides    EK,  FH   of  the   gnomon  are   given 
but  EK  is  equal  to    DC,  and  FH  to   DB  ;  vdierefore    CD,  DBe82.dat. 
j^rc  each  of  them  given. 

3  K 


c  5(^.  dat. 

d  36.  and 
43.  1. 


442  EUCLID'S 

This  demonstration  is  the  analysis  of  the  problem  in  the  J8tli 
prop,  of  book  6,  the  construction  and  demonstration  of  which 
proposition  is  the  composition  of  the  analysis  ;  and  because  the 
given  space  AC  or  its  equal  the  gnomon  ELH  is  to  be  taken  from 
the  figure  EF  described  upon  the  half  of  AB  similar  to  BC, 
therefore  AC  must  not  be  greater  than  EF,  as  is  shovn  in  the 
27th  prop.  B.  6. 


59. 


PROP.  LXXXIV. 


IF  a  parallelogram  equal  to  a  given  space  be  applied 
to  a  given  straight  line,  exceeding  by  a  parallciogram 
given  in  species ;  the  sides  of  the  excess  are  given. 


G       ] 

F   H 

A 

E 

\ 

Let  the  parallelogram  AC  equal  to  a  given  space  be  applied 
to  the  given  straight  line  AB,  exceeding  by  the  parallelogram 
BDGL  given  in  species  ;  each  of  the  straight  lines  CD,  DB  are 
given. 

Bisect  AB  in  E  ;  therefore  EB  is  given  in  magnitude ;  upon 

a  18. 6.     EB  d(:scribe=i  the  parallelogram  EF  similar  to  LD,  and  similar- 
ly placed  ;    therefore  EF  is  given  in    species,  and   is   about  the 

b  26.  6.  same  diameter  *»  with  LD.  Let  CBG  be 
the  diameter  and  construct  the  figure  : 
therefore,  because   the  figure   EF  given 

in   species   is  described   upon  the  given      A.      ^_, IbJii_jD 

straight  line  EB,  EF  is  given  in  magni- 

c  56.clat.  tude  c,  and   the    gnomon  ELH  is    equal 

d  36.  and  to    the    given  figure  '^    AC  ;    wherefore, 

43.  1.       siiice  EF  is  encreased  by  the  given  gnomon  ELH,  its  sides  EK 

e  82.  dat.  I'^i  ^''*^  given  «  ;  but  EK  is  equal  to  CD,  and  FH  to  BD ;  there- 
fore CD,  DB  are  each  of  them  given. 

Tbis  demonstration  is  the  analysis  of  the  problem  in  the  29th 
prop,  book  6,  the  construction  and  demonstration  of  which  is 
the  composition  of  the  analysis. 

CoK.  If  a  parallelogram  given  in  species  be  applied  to  a  given 
stnj:;ht  line,  exceeding  by  a  parallelogram  equal  to  a  given 
space  ;  the  sides  of  the  parallelogram  are  given. 

Let  the  pardlle!o;.^ram  ADCE  given  in  species  be  applied  to 
the  given  straight  line  AB  exceeding  by  tHe  parallelogram  BDCG 
equjil  to  a  given  space  ;  the  sides  AD,  DC  of  the  parallelogram 
are  given. 


K      L    C 


\ 

r 

^ 

K 


A  B      D 


DATA.  U3 

Draw  the  diameter  DE  of  the  parallelogram  AC,  and  con- 
struct the  figure.     Because  the   parallelogram  AK  is  equal=*  to*^'^- ■^• 
BC    which  is  given,  therefore    AK    is         E  G       C  b24  6 

given  ;  and  BK  is  similar^  to  AC,  there- 
fore BK  is  given  in  species.     And  since 
the  parallelogram  AK  given  in  magni-      „ 
tude  is  applied  to  the  given  straight  line 
AB,  exceeding  by  the  parallelogi-am  BK 
given  in  species,  therefore  by  this  pro- 
position, BD,  DK  the  sides   of  the   excess   are   given,  and    the 
straight  line  AB  is  given ;  therefore  the  whole  AD,  as  also  DC, 
to  which  it  has  a  given  ratio,  is  given. 

PROB. 

To  apply  a  parallelogram  similar  to  a  given  one,  to  a  given 
straight  line  AB,  exceeding  by  a  parallelogram  equal  to  a  given' 
space. 

To  the   given  straight  line  AB  apply  <=  the  parallelogram  AK  =  29.  £ 
equal  to  the  given   space,  exceeding   by  the   parallelogram  BK 
similar  to  the  one  given.     Draw  DF    the  diameter  of  BK,  and 
through  the  point  A  draw  AE  parallel  to  BF  meeting  DF  pro- 
duced in  E,  and  complete  the  parallelogram  AC. 

The  parallelogram  BC  is  equal »  to  AK,  that  is,  to  the  given 
space  ;  and  the  parallelogram  AC  is  similar  ^  to  BK  ;  therefore 
the  parallelogram  AC  is  applied  to  the  straight  line  AB  similar 
to  the  one  given  and  exceeding  by  the  parallelogram  BC  which 
is  equal  to  the  given  space. 


PROP.  LXXXV.  84. 

IF  two  straight  lines  contain  a  parallelogram  given 
in  magnitiide,  in  a  given  angle ;  if  the  difference  of 
the  straight  lines  be  given,  they  shall  each  of  them  be 
given. 

Let  AB,  BC  contain  the  parallelogram  AC  given  in  magni- 
tude, in  the  given  angle  ABC,  and  let  the  excess  of  BC  above 
AB  be  given  ;  each  of  the  straight  lines  AB.  BC  is  given. 

Let  DC  be  the  given  excess   of  BC  above  A  E 

BA,  therefore  the  remainder  BD  is  equal 
to  BA.  Complete  the  parallelogram  AD  ; 
and  because    AB   is  equal   to  BD,  the  ratio 

of  ABto  BD  is  given  ;    and  the  angle  ABD      ' :^ — ^ 

is  given,  therefore  the  parallelogram    AD  is 

given  in  species;  and    because   the  given  parallelogram  AC   is 


4'U  -  EUCLID'S 

applied  to  the  given  straight  line  DC,  exceeding  by  the  paral- 
lelogram AD  given  in  specitis,  the  sides  of  the  excess  are  gi- 
j,84.  dat.  ven  *:  therefore  BD  is  given  ;  and  DC  is  given,  wherefore  me 
whole  BC  is  given  :  and  AB  is  given,  therefore  AB,  BC  arc  each 
of  them  given. 


85.  PROP.  LXXXVI. 

IF  tv;0  straight  lines  contain  a  parallelogram  gi\cn  in 
magnitude,  in  a  given  angle  ;  if  both  of  them  together 
be  gi\'en,  they  sliall  each  of  them  be  given. 

Let  the  two  straight  lines  AB,  BC  contain  the  parallelogi'am 
AC  given  in  magnitude,  in  the  given  angle  ABC,  and  kt  \B, 
BC  together  be  given  ;  each  of  tlie  straight  lines  AB,  BC  is 
given. 

Produce  CB,  and  make  BD  equal  to  AB,  and  complete  the 
parallelogram  ABDE.  Because  DB  is  equal  to  BA,  ana  the 
angle  ABD  given,  because  the  adjacent  an-         E  A 

gle  ABC  is  given,  the  parallelogram  AD  is 
given  in  species :  and  because  AB,  BC  to- 
gether are  given,  and  AB  is  equal  to  BD  ; 
therefore  DC  is  given  :  and  because  the  gi- 
ven parallelogram  AC  is  applied  to  the  gi- 
ven straight  line  DC,  deficient  by  the  parallelogram  AD  given 
a83.  dat,!"^  species,  the  sides  AB,  BD  of  the  defect  are  given  =*  ;  and  DC 
is  given,  wherefore  the  remainder  BC  is  given  ;  and  each  of  the 
straight  lines  AB,  BC  is  therefore  given. 


87.  (  PROP.  LXXXVI  I. 

IF  t^^•o  straight  lines  contain  a  parallelogram  gi\-en  in 
magnitude,  in  a  given  angle ;  if  the  excess  of  the  square 
of  the  greater  above  the  square  of  the  lesser  be  gnen, 
each  of  the  straight  lines  shall  be  gi\en.' 

Let  the  two  straight  lines  AB,  BC  contain  the  given  parallelo- 
gram AC  in  the  given  angle  ABC  ;  if  the  excess  of  the  squaie 
of  BC  above  the  square  of  BA  be  given  ;  AB  and  BC  arc  each 
of  them  given. 

Let  tlie  given  excess  of  the  square  of  BC  above  the  square 
of  BA  be  the  rectangle    CB,  BD  :  take  this   from   the  square 


DATA.  445 

of  BC,  the  remainder,  which  is  *  the  rectangle    BC,  CD  is  e»a2.  2. 
qu,.l  to  tlie  square    of    AB  ;    and  because  the  angle    ABC    of 
Iht    purullelogram   AC  is  given,  the  ratio  of   the  rectangle    of 
the  sides  AB,  BC  to  the  parallelogram  AC  is  given  ^  ;    and  AC  b  62.  dar. 
is   ii,ivcn,   therefore   the    rectangle  AB,  BC  is  given  ;    and  the 
rectangle  CB,  BD  is  given  ;   therefore  the  ratio  of  the  rectangle 
CB,    BD   to   the  rectangle  AB,  BC,  that  is':,  the  ratio  of   thee  1.6. 
straight  line  DB  to  BA  is  given:  therefore 'I  the    ratio   of  the  d  54.  tht." 
square  of  DB  to  the  square  of  BA  is  given,         A 
and  the  square  of  BA  is  equal  to  the  rect- 
angle BC'.  CD:    wherefore  the  ratio  of  the 
rectangle  BC,  CD  to  the  square  of   BD  is 
given,  as  also  the  ratio  of  four    times    the  B  PD  C 

rectangle  BC-  CD  to  the  square  of  BU  ;  and, 

by  composition  e,  the  ratio  of  four  times  the  i-^ctangle  BC,  CDeT.  dat, 
together  with  the  square  of    BC  to  the  square  of  BD  is  given  : 
but  four  times  the  rectangle  BC,  CD  together  with  the  square 
of    BD  is  equal  f    to  the  square  of   the  straight  lines  BC,  CD^S.  2. 
taken  together:    therefore  the  ratio  of   the    square  of    BC,    CD 
together  to  tlie  square  of  liD  is  given  ;   wherefores  the  ratio  ofS^^.dat, 
the  straiglu  line  BC  together  with  CD  to  BD  is  given  :  and,  by 
composition,  the  riitio  of    BC  together  with  CD  and  DB,  that 
is,  the  ratio  of  twice  BC  to  BD,  is  given  ;  therefore  the  ratio  of 
BC  to  BD  is  given,  as  also<=  the  ratio  af  the  square  of  BC  to  the 
rectangle  CB,  BD  :   but  the  rect^ingle  CB,  BD  is  given,  being 
the    given   excess  of  the   squares  of    BC,    BA  ;    therefore   the 
square  of   BC,  and  the  straight  line  BC  is  given  :  and  the  ratio 
of  BC  to  BD,  as  also  of  BD  lo  BA  has  been  shown  to  be  given  ; 
therefore  ii  the  ratio  of  BC  to  BA  is  given  ;    and  BC  is   given,  h  9.  da', 
wherefore  Bi^  is  given. 

Tlie  preceding  demonstration  is  the  analysis  of  this  problem, 
viz. 

A  parallelogram  AC  which  has  a  given  angle  ABC  being  gi- 
ven in  magnitude,  and  the  excess  of  the  squt.re  of  hC  one  of  its 
sides  above  the  square  of  the  olher  BA  being  given  ;  to  fmd  the 
sides:  and  the  composition  is  as  follows. 

Let  EFG  be  the  given  angle  lo  which  the  angle  ABC  is  re- 
quired to  be  equal,  and  from  aiiy  point  E  in  i'"E,  araw  ECi 
perpendicular  to  FG  ;  let  the  rect- 
angle EG,  GH  be  the  given  space 
to  which  the  paralielogran^  AC  is 
to  be  made  equal  ;  auU  the  rectan- 
gle HG,  GL,  be  tlie  givtn  excess 
of  the  squares  of  BC,  3A. 

Takv,  in  the  straight  line  GE, 
GK  equal  to  FE,  and  make  GM        F     G       L     O  H  N 

double  of  GK :  join  AIL,  and  in  GL  produced,  take  LN  equal  to 
LM  :  bisect  CN  in  O..  and  between  CfL  GO  {ind  a  mean  pro- 


446 


EUCLID'S 


portional  BC  :  as  OG  to  GL,  so  make  CB  to  BD ;  and  malie  the 
angle  CBA  equal  to  Gr  E,  and  as  LG  to  GK  so  make  DB  to  B A  ; 
and  complete  the  parallelogram  AC :  AC  is  equal  to  the  rect.. 
angle  EG,  GH,  and  the  excess  of  the  squares  of  CB,  BA  is  equal 
to  the  rectangle  HG,  GL. 

Because   as  CB  to  BD,  so  is  OG  to  GL,  the  square  of  CB 

J 1. 6.  is  to  the  rectangle  CB,  BD  as  *  the  rectangle  HG,  GO  to 
the  rectangle  HG,  GL  :  and  the  square  of  CB  is  equal  to  the 
rectangle  HG,  GO,  because  GO,    BC,  GH  are    proportionals ; 

b  14.  5.  therefore  the  rectangle  CB,  BD  is  equal  ^  to  HG,  GL.  And 
because  as  CB  to  BD,  so  is  OG  to  GL ;  twice  CB  is  to  BD, 
as  twice  OG,  that  is,  GN  to  GL  ;  and,  by  division,  as  BC 
together  with  CD  is  to   BD,  so  is   NL,  that   is,   LM,  to  LG ; 

c22.  6.  therefore <=  the  square  of  BC  together  with  CD  is  to  the  square 
of   BD,    as   the  square  of   ML  to  the  square  of  LG  :    but  the 

d  8. 2.  square  of  BC  and  CD  together  is  equal  <1  to  four  times  the 
rectangle  BC,  CD  together  with  the  square  of  BD :  therefore 
four  times  the  rectangle  BC,  CD  together  with  the  square  of 
BD  is  to  the  square  of  BD,  as  the  square  of  ML  to  the  square 
of  LG  :  and,  by  division,  four  times  the  rectangle  BC,  CD  is 
to  the  square  of  BD,  as  the  square  of  MG  to  the  square  of 
GL  ;  wherefore  the  rectangle  BC,  CD  is  to  the  square  of  BD 
as  (the  square  of  KG  the  half  of  MG  to  the  square  of  GL» 
that  is,  as)  the  square  of  AB  to  the  square  of  BD,  because  as 
LG  to  GK,  so  DB  was  made  to  BA  :  therefore i'  the  rectan- 
gle BC,  CD  is  equal  to  the  square  of  AB.  To  each  of  these  add 
the  rectangle  CB,  BD,  and  the  square  of  BC  becomes  equal 
to  the  square  of  AB  together  with  the  rectangle  CB,  BD  ; 
therefore  this  rectangle,  that  is,  the  given  rectangle  HG,  GL, 
is  the  excess  of  the  squares  of  BC,  AB.  From  the  point  A, 
draw  AP  perpendicular  to  BC,  and  hrcause  the  angle  ABP 
is  equal  to  the  angle  EFG  the  triangle  ABP  is  equiangular 
to  EFG:  and  DB  was  made  to  BA,  as  LG  to  GK  ;  therefore 
as  the  rectangle  CB,  BD  to  CB,  BA,  so  is  the  rectangle  HG, 

M 


PD 


F    G 


O 


HN 


GL  to  HG,    GK;  and  as  the  rectangle  CB,    BA  to  AP,    BC, 
so  is   (the    straight'  line  BA  to  AP,   and    so  is  FE  or  GK  ta 


DATA.  447 

EG,  and  so  is)  the  rectangle  HG,  GK  to  HG,  GE ;  therefore 
ex  xqualii  as  the  rectangle  CB,  BD  to  AP,  BC,  so  is  the  rect- 
angle HG,  GL  to  EG,  GH:  and  the  rectangle  CB,  BD  is  equal 
to  HG,  GL ;  therefore  the  rectangle  AP,  BC,  that  is,  the  paral- 
lelogram AC,  is  equal  to  the  given  rectangle  EG,  GH. 


PROP.  LXXXVIII.  n;. 


IF  two  straight  lines  contain  a  parallelogram  given 
in  magnitude,  in  a  given  angle ;  if  the  sum  of  the 
squiires  of  its  sides  be  given,  the  sides  shall  each  of  them 
be  given. 


Let  the  two  straight  lines  AB,  BC  contain  the  parallelogram 
ABCD  given  in  magnitude  in  the  given  angle  ABC,  and  let  the 
sum  of  the  squares  of  AB,  BC  be  given  ;  AB,  BC  are  each  of 
them  given. 

First,  let  ABC  be  a  right  angle  ;  and  because  twice  the  rect- 
angle contained  by  two  equal  straight  lines  is  equal  to  both 
their  squares  ;  but  if  two  straight  lines  are  un-     A  D 

equal,  twice  the  rectangle  contained  by  them  is        l  J 

less  than  the  sum  of  their  squares,  as  is  evident         I 

from  the  7th  prop,  book  2,  Elem  ;  therefore  twice     B  C 

the  given  space,  to  which  space  the  rectangle  of  which  the  sides 
are  to  be  found  is  equal,  must  not  be  greater  than  the  given  sum 
of  the  squares  of  the  sides :  and  if  twice  that  space  bt;  equal  to 
the  given  sum  of  the  squares,  the  sides  of  the  rectangle  must  ne- 
cessarily be  equi*l  to  one  another  ;  therefore  in  this  case  describe 
a  square  ABCD  equal  to  the  given  rectangle,  and  its  sides  AB, 
BC  are  those  which  were  to  be  found :  for  the  rectangle  AC  is 
equal  to  the  given  space,  and  the  sum  of  the  squares  of  its  sides 
AB,  BC  is  equal  to  twice  the  rectangle  AC,  that  is,  by  the  hypo- 
thesis, to  the  given  space  to  which  the  sum  of  the  squares  was 
required  to  be  equal. 

But  if  twice  the  given  rectangle  be  not  equal  to  the  given  sum 
of  the  squares  of  the  sides,  it  must  be  less  than  it,  as  has  been 
shown.      Let  ABCD  be  the  rectangle,  joi'n  AC  and  draw  BE 


*i8  EUCLID'S 

perpendicular  lo  it.  and  complete  the  rectangle  AEEi',  and  de-- 
scribe  the  circle  ABC  about  the  triai  gle  i\EC;  AC  is  its  duimc- 

aCor.5.4.  tgj,  a  ;  and  because  the  triangle  ABC  is  similar  ^  to  AEB,  as  AC  to 

^^•^-  CB  so  is  AB  to  BE  ;  thercibre  the  rectangle  AC,  BE  is  <.qt;ul  to 
AB,  BC  ;  and  the  rectangle  AB,  BC  is  given,  \vherLIbr^J  .iC, 
BE  is  given:   and  because  t!^e  sum  of  the  squares  of  aB,  B-^    is 

c  4r.  1.  given,  the  square  of  AC  which  is  equal  ^  to  that  sum  is  given  ;  and 
AC  itself  is  therefore  given  in  magnitudi^  :  let  AC  be  likewise  given 

d32.  dat. in  position,  and  the  jjoint  A;  therefore  AF  is  given  '^  in  position  : 
and  the  rectangle  AC,  BE  is  given,  as  lias 

e  61.dat.  been  shown,  and  7\.C  is  given,  vvhei'cfore  « 
BE   is  given  in  magnitude,  as  also  AF 
which  is  equal  to  it ;  and  AF  is  also  giv-  ,^  ^  Ij 
en  in  position,  and  the  point  A  is  given  ;      ^^V 

f  30.  dat.  wherefore  f  the  point  F  is  given,  and  the 

g31.  dat.  straight  line  FB  in  position  a  :  and  the  cir- 
cumference   ABC   is    given    in   position, 

li28.  dat.  wherefore  ^  the  point  B  is  given  :  and  the     G         K 

points  A,  C  are  given  ;  therefore  the  straight  lines  AB,  BC  are. 

i29.dat.  given  >  in  position  and  magnitude. 

The  sides  AB,  BC  of  the  rectangle  may  be  found  thus  ;  let 
the  rectangle  GH,  GK  be  the  given  space  to  which  the  rect- 
angle AB,  BC  is  equal  ;  and  let  GH,  GL  be  the  given  rect- 
angle  to   which  the  sum  of  the   squares  of  A3,  BC  is  equal : 

k  14.  2.  ^-^^^  ^  ^  square  equal  to  the  rectangle  GH,  GL  :  and  let  its  side 
AC  be  given  in  position  ;  upon  AC  as  a  diameter  describe  the 
semicircle  ABC,  and  as  AC  to  GH,  so  make  GK  to  AF,  and  from 
the  point  A  place  AF  at  right  angles  to  AC :  therefore  the  rect- 

ri6. 6.  angle  CA,  AF  is  equaP  to  GH,  GK;  and,  by  the  hypothesis, 
twice  the  rectangle  GH,  GK  is  less  than  GH,  GL,  that  is,  than 
the  square  of  AC;  wherefore  twice  the  rectangle  C.\,  AF  is  less 
than  the  square  of  AC,  and  the  rectangle  CA,  AF  itself  less 
than  half  the  square  of  AC,  that  is,  than  the  rectangle  contain- 
ed by  the  diameter  AC  and  its  half ;  wherefore  AF  is  less  than 
the  semidiameter  of  the  circle,  and  consequently  the  straight  line 
drawn  through  the  point  F  parallel  to  AC  must  meet  the  circum- 
ference in  two  points:  let  B  be  cither  of  them,  and  joint  AB, 
BC,  and  complete  the  rectangle  ABCD,  ABCD  is  the  rectangle 
which  was  to  be  found:  draw  BE  perpendicular  to  AC;  there- 
in 34.  1.  fo'"'-'  3E  is  equal '"  to  AF,  and  because  the  angle  ABC  in  a  semi- 
circle is  a  right  angle,  the  rectangle  AB,  BC  is  equal  ^  to  AC, 
BE,  that  is,  to  the  rectangle  CA,  AF,  which  is  equal  to  the  given 
rectangle  GH,  GK  ;  and  the  squares  of  AB,  BC  are  together 
equal  <^  to  the  square  of  AC,  that  is,  to  the  given  rectangle  GH, 
.GL. 


DATA. 


449 


B  L 


But  if  the  given  angle  ABC  of  the  parallelogram  AC  be  not  a 
right  angle,  in  this  case,  because  ABC  is  a  given  angle,  the  ra- 
tio of  the  rectangle  contained  by  the  sides  AB,  BC  to  the  paral- 
lelogram AC  is  given";  and  AC  is  given,  therefore  the  rcctan-n62.dat. 
gle  AB,  BC  is  given  ;  and  the  sum  of  the  squares  of  AB,  BC  is 
given  ;  therefore  the  sides  AB,  BC  are  given  by  the  preceding 
case. 

The  sides  AB,  BC  and  the  parallelogram  AC  may  be  found 
thus :  let  EFG  be  the  given  angle  of  the  parallelogram,  and 
from  any  point  E  in  FE  draw  EG  perpendicular  to  FG ;  and 
let  the  rectangle  EG,  FH  be  the  given  space  to  which  the  pa- 
rallelogram is  to  be  made  equal,  and  let  EF,  A  D 
FK  be  the  given  rectangle  to  which  the 
sum  of  the  squares  of  the  sides  is  to  be  equal. 
And,  by  the  preceding  case,  find  the  sides 
of  a  rectangle  which  is  equal  to  the  given 
rectangle  EF,  FH,  and  the  squares  of  the 
sides  of  which  are  together  equal  to  the  gi- 
ven rectangle  EF,  FK;  therefore,  as  was 
shown  in  that  case,  twice  the  rectangle  EF, 
FH  must  not  be  greater  than  the  rectangle 
EF,  FK ;  let  it  be  so,  and  let  AB,  BC  be 
the  sides  of  the  rectangle  joined  in  the  an- 
gle ABC  equal  to  the  given   angle    EFG, 

and  complete  the  parallelogram  ABCD,  which  will  be  that  which 
was  to  be  found;  draw  Ai>  perpendicular  to  BC, and  because  the 
angle  ABL  is  equal  to  EFG,  the  triangle  ABL  is  equiangular  to 
EFG ;  and  the  parallelogram  AC,  that  is,  the  rectangle  AL,  BC 
is  to  the  rectangle  AB,  BC  as  (the  straight  line  AL  to  AB,  that 
is,  as  EG  to  EF,  that  is,  as)  the  rectangle  EG,  FH  to  EF,  FH  -, 
and,  by  the  construction,  the  rectangle  AB,  BC  is  equal  to  EF, 
FH,  therefore  the  rectangle  AL,  BC,  or,  its  equal,  the  paralle- 
logram AC,  is  equal  to  the  given  rectangle  EG,  FH  ;  and  the 
squares  of  AB,  BC  are  together  equal,  by  construction,  to  the 
given  rectangle  EF,  FK. 


L 


450  EUCLID'S 


86,  PROP.  LXXXIX. 


IF  two  straight  lines  contain  a  given  parallelogram  in  a 
given  angle,  and  if  the  excess  of  the  square  of  one  of  them 
above  a  given  space,  has  a  given  ratio  to  the  square  of  the 
other ;  each  of  the  straight  lines  shall  be  given. 


Let  the  two  straight  lines  AB,  BC  contain  the  given  parallelo- 
gram AC  in  the  p;iven  angle  ABC,  and  let  the  excess  of  the 
square  of  BC  above  a  given  space  have  a  given  ratio  to  the  square 
©f  AB,  each  of  the  straight  lines  AB,  BC  is  given. 

Because  the  excess  of  the  square  of  BC  above  a  given  space 
lias  a  given  ratio  to  the  square  of  BA,  let  the  rectangle  CB,  BD 
be  the  given  space  ;  take  this  from  the  square  of  BC,  the  re- 
a  2.  2.     mainder,  to  wit,  the  rectangle  =^  BC,  CD   has   a  given  ratio  to 
the  square  of  BA  :  draw  AE  perpendicular  to  BC,  and  let  the 
square  of  BF  be  equal  to  the  rectangle  BC,  CD,  then,  because 
the  angle  ABC,  as  also  BEA,  is  given,  the 
b  43  dat.  triangle  ABE  is  given  ^^  in  species,  and  the 
ratio  of  AE  to  AB  given  :    and  because  the 
ratio  of   the  rectangle  BC,   CD,    that  is,   of 
the  square  of  BF  to  the  square  of  BA,  is  gi- 
ven, the  ratio  of  the  straight  line  BF  to  BA 
c  58.  dat.  is  given  «  ;  and  the  ratio  of  AE  to  AB  is  given,  wherefore  ^Lthe 
d9,  dat.  ratio  of  AE  to  BF  is  given  ;  as  also  the  ratio  of  the  rectangle 
e  35. 1.    AE,  BC,  that  is,  c  of  the  parallelogram  AC  to  the  rectangle  KB, 
BC  ;  and  AC  is  given,  wherefore  the  rectangle  FB,  BC  is  given. 
The  excess  of  thg  square  of  BC  above  the  square  of  BF,  that 
is,  above  the  rectangle  BC,  CD,  is  given,  for  it  is  equal  ^  to  the 
given  rectangle  CB,  BD  ;  therefore,  because  the  rectangle  con- 
tained by  the  straight  lines  FB,  BC  is  given,  and  also  the  excess 
of  the  square  of  BC  above  the  square  of  BF ;  FB,  BC  are  each 
(87-.  clat  of  them  given  ^ ;  and  the  ratio  of  FB  to  BA  is  given  ;  therefore, 
AB,  BC  are  given. 


The  cornpos'Uion  is  as  follows  ; 

Let  CilK  be  the  given  siigle  to  which  the  angle  of  the  paral- 
lelogram is  to  be  made  equal,  and  from  any  point  G  in  IIG,  draw 
CK.  perpendicular  to  HK  j  let  GK,  IIL  be  the  rectangle  to  which 


DATA. 


4:5  i 


H  KM 


A 


7 


/ 


c 


the  parallelogram  is  to  be  made  equal,  and  N 

let  LH,  HM  be  the  rectangle  equal  to  the 

giten  space  which  is  to  be  taken  from  the 

square  of  one  of  the  sides;  and  let  the  ratio 

of  the  remainder  to  the  square  of  the,  other 

side  be  the  same  with  the  ratio  of  the  square 

of  the  given  straight  line  NH  to  the  square  of  the  given  straight 

line  HG. 

By  help  of  the  87th  dat.  find  two  straight  lines  BC,  BF,  which 
contain  a  rectangle  equal  to  the  given  rectangle  NH,  HL,  and 
such  that  the  excess  of  the  square  of  BC  F 

above  the  square  of  BF  be  equal  to  the  gi- 
ven rectangle  LH,  HM ;  and  join  CB,  BF 
in  the  angle  FBC  equal  to  the  given  angle 
GHK:  and  as  NH  to  HG,  so  make  FB  to 
BA,  and  complete  the  parallelogram  AC,  BED 
and  draw  AE  perpendicular  to  BC  ;  then 

AC  is  equal  to  the  rectangle  f  K,  IIL  ;  and  if  from  the  square 
of  BC,  the  given  rectangle  LH,  HM  be  taken,  the  remainder 
shall  have  to  the  square  of  BA  the  same  ratio  which  the  square 
of  NH  has  to  the  square  of  HG. 

Because,  by  the  construction,  the  square  of  BC  is  equal  to  the 
square  of  BF,  together  with  the  rectangle  LH,  HM  ;  if  from  the 
square  of  BC  there  be  taken  the  rectangle  LH,  HM,  there  re- 
mains the  square  of  BF  which  has  s  to  the  square  of  BA  the  same  „  22.  6. 
ratio  vvhich  the  squai'e  of  NH  has  to  the  square  of  HG,  because, 
as  NH  to  HG,  so  FB  was  made  to  BA ;  but  as  HG  to  GK,  so  is 
BA.  to  AE,  because  the  triangle  GHK  is  equiangular  to  ABE; 
therefore,  ex  aguali,  as  NH  to  GK,  so  is  FB  to  AF  ;  wherefore  ^^h  1.  Gf 
the  rectangle  NH,  HL  is  to  the  rectangle  GK,  HL,  as  the  rect- 
angle FB,  BC  to  AE,  BC  ;  but  by  the  construction,  the  rectangle 
NH,  HL  is  equal  to  FB,  BC  ;  therefore  '  the  rectangle  GK,  HL  i  14.  5: 
is  equal  to  the  rectangle  AE,  BC,  that  is,  to  the  parallelogram 
AC. 

The  analysis  of  this  problem  might  have  been  made  as  In  the 
86th  prop,  in  the  Greek,  and  the  composition  of  it  may  be  made 
as  that  which  is  in  prop.  87th  of  this  edition. 


45C  EUCLID'S 


0,  PROP.  XC. 

IF  hvo  straight  lines  contain  a  given  pai'alielogram  in 
a  given  angle,  and  if  the  square  of  one  of  them  together 
with  the  space  Vvhich  has  a  gi\'cn  ratio  to  the  squai'e  of 
the  other  be  given,  each  of  the  straight  lines  shall  be  given. 

Let  the  two  straight  lines  AB,  BC  contain  the  given  parallelo- 
gram AC  in  the  given  angle  ABC,  and  let  tlie  square  of  BC  to- 
gether with  the  space  which  has  a  given  ratio  to  the  square  of  AB 
be  given,  AB,  BC  are  each  of  them  given. 

Let  the  square  of  BD  be  the  space  which  has  the  given  ratio 
to  the  square  of  AB  ;  therefore,  by  the  hypothesis,  the  square 
of  BC  together  with  the  square  of  BD  is  given.     From  the  point 
A,  draw  AL  perpendicular  to  BC  ;  and  because  the  angles  ABE, 
a43.dat.  BE  A  are  given,  the  triangle  ABE  is  given  »  in   species;  there- 
fore the  ratio  of  BA  to  AE  is  given  :  and  because  the  ratio  of 
the  square  of  BD  to  the  square  of  BA  is  given,  the  ratio  of  the 
b  58  dat.  straight  line  BD  to  B  A  is  given  ^ ;  and  the  ratio  of  BA  to  AE 
c  9.  dat    is  given  ;  therefore  ^  the  ratio  of  AE  to  BD  is  given,  as  also  the 
ratio  of  the  reciangle  AE,  BC,  that  is,  of  the  parallelogram  AC 
to  the  rectangle  DB,  BC  ;  and  AC  is  given,  therefore  the  rect- 
angle DB,  BC  is  given  ;  and  the  square  of  BC  together  with  the 
D  M 

A 


"7 
y 


BE  C 


GH  K 


d88.dat  square  of  BD  is  given  ;  therefore  ^  because  the  rectangle  contain- 
ed by  the  two  straight  lines  DB,  BC  is  given,  and  the  sum  of 
their  squares  is  given  :  the  sti-aight  lines  DB,  BC  are  each  of 
them  given  ;  and  the  ratio  of  DB  to  BA  is  given  ;  therefore  AB, 
BC  are  given. 


The  comflosiiion  is  as  fellows : 

Let  FGH  be  the  given  angle  to  vvhich  the  angle  of  the  paral- 
lelogram is  to  be  made  equal,  and  from  any  point  F  in  OF 
draw  FH  perpendicular  to  GH  ;  and  let  the  rectangle  FH, 
GK  be  tliat  to  which  the  parallelogram  is  to  be  made  equal; 
and  let  the  rectangle  KG,  GL  be  the  space  to  which  the  square 


DATA. 


453 


of  one  of  the  sides  of  the  parallelogram  together  with  the  space 
which  has  a  given  ratio  to  the  square  of  the  other  side,  is  to  be 
made  equal  ;  and  let  this  given  ratio  be  the  same  which  the  square 
of  the  given  straight  line  MG  has  to  the  square  of  GF. 

By  the  88th  dat.  find  two  straight  lines  DB,  BG  which  con- 
tain a  rectangle  equal  to  the  given  rectangle  MG,  GK,  and 
such  that  the  sum  of  their  squares  is  equal  to  the  given  rectan- 
gle KG,  GL :  therefore,  by  the  determination  of  the  pro- 
blem in  that  proposition,  twice  the  rectangle  MG,  GK  must 
not  be  greater  than  the  rectangle  KG,  GL.  Let  it  be  so,  and 
join  the  straight  lines  DB,  BC  in  the  angle  DEC  equal  to  the 
given  angle  FGH  ;  and,  as  MG  to  GF,  so  make  DB  to  BA, 
and  complete  the  parallelogram  AC  :   AC  is  equal  to  the  rect- 


D 


M 


BE 


7   a 


K 


angle  FH,  GK;  and  the  square  of  BC  together  with  the  square 
of  BD,  which,  by  the  construction,  has  to  the  square  of  BA  the 
given  ratio  which  the  square  of  MG  has  to  the  square  of  GF,  is 
equal,  by  the  construction,  to  the  given  rectangle  KG,  GL. 
Draw  AE  perpendicular  to  BC. 

Because,  as  DB  to  BA,  so  is  MG  to  GF ;  and  as  BA  to  AE, 
so  GF  to  FH  ;  ex  aguali,  as  DB  to  AE,  so  is  MG  to  FH  ;  there- 
fore, as  the  rectangle  DB,  BC  to  AE,  BC,  so  is  the  rectangle 
MG,  GK  to  FH,  GK  ;  and  the  rectangle  DB,  BC  is  equal  to  the 
rectangle  MG,  GK  ;  therefore  the  rectangle  AE,  BC,^  that  is, 
the  parallelogram  AC,  is  equal  to  the  rectangle  FH,  GK. 


PROP.  XCI. 


88. 


IF  a  straight  line  drawn  vrithin  a  circle  given  in 
magnitude  cuts  off  a  segment  Vvbich  contains  a  given 
angle  ;  the  straight  line  is  given  in  magnitude. 

In  the  circle  ABC  given  in  magnitude,  let  the  straight  line 
AC  be  drawn,  cutting"  off  the  segment  AEC  which  contains  the 
given  angle  AEC  ;  the  straight  line  AC  is  given  in  magnitude. 
"  Take  D  the  centre  of  the  circle  ^   join  AD  and  produce  Hal.  3. 


■454 


EUCLID'S 


to  E,  and  join  EC :  the  angle  ACE  being 

b  ol.  3.   a  right  ^  angle  is  given  ;    and  the  angle 

>c4j.  dat.  AEC  is  given  ;  therefore  =  the  triangle  ACE 

is  given  in  species,  and  the  ratio  of  EA  to 

AC  is  therefore  given  ;  and  EA  is  given  in 
d  5.  def.  magnitude,  because  the  circle  is  given  ^  in 
e  a.  dat.  magnitude;  AC  is  therefore  given  ^  inmag« 

nitude. 


89.  PROP.  XCII. 

IF  a  straight  line  giA^en  in  magnitude  be  drawn 
within  a  circle  given  in  magnitude,  it  shall  cut  off  a 
segment   containing  a  given  angle. 

Let  the  straight  line  AC  given  in  magnitude  be  drawn  within 
the  circle  ABC  given  in  magnitude  ;  it  shall  cut  off  a  segment 
contaning  a  given  angle. 

Take  D  the  centre  of  the  circle,  join  AD 
and  produce  it  to  E,  and  join  EC :  and  be- 
cause each  of  the  straight  lines  EA  and  AC 

a  1.  dat.  is  given,  their  ratio  is  given  ^ ;  and  the  an- 
gle ACE  is  a  right  angle,  therefore  the  tri- 

b46.dat.  angle  ACE  is  given ''  in  species,  and  conse- 
quently the  angle  AEC  is  given. 


90. 


PROP.  XCIIL 


IF  from  any  point  in  the  circumference  of  a  circle 
given  in  position  two  straight  lines  be  drawn  meet- 
ing the  circumference  and  containing  a  given  angle ; 
if  the  point  in  w^hich  one  of  them  meets  the  circum- 
ference again  be  given,  the  point  in  which  the  other 
meets  it  is  also   given. 

From  any  point  A  in  the   circumference  of  a  circle  ABC  gi- 
ven in  position,  let  AB,  AC  be  drawn  to  the  circumference,  ma- 
king the  given  angle  BAC  ;  if  the  point  B 
be  given,  the  point  C  is  also  given. 

Take    D    the   centre  of  the  circle,    and 
join  BD,  DC  ;  and  because  each   of  the 
a  29.  dat. points  B,  D  is  given,  BD  is  given  ^^  in  po- 
sition ;  and  because  the  angle  BAC  is  gi-  B 
b  20.  3.   ven,  the  angle  BDC  is  given  ^,  therefore 


DATA* 


455 


because  tlie   straight  line  DC  is  drawn  to  the  given  point  D  in 

the  straight  hne  BD  given  in  position  in  the  given  angle  BDC, 

DC.  is  given  <=  in  position  :  and  the  circumference  ABC  is  given  c  32.  dat. 

in  position,  therefore^  the  point  C  is  given.  d28.dat. 


PROP.  XCIV.  ^* 

If  from  a  given  point  a  straight  line  be  drawn  touch- 
ing a  circle  given  in  position  ;  the  straight  line  is  given 
in  position  and  magnitude. 

Let  the  straight  line  AB  be  drawn  from  the  given  point  A 
touching  the  circle  BC  given  in  position  ;  AB  is  given  in  posi- 
tion and  magnitude. 

Take  D  the  centre  of  the  circle,  and  join  DA, 
each  of  the  points  D,  A  is  given,  the 
straight  line  AD  is  given  *  in  position 
and  magnitude  :  and  DBA  is  a  right'' 
angle,  wherefore  DA  is  a  diameter*  of 
the  circle  DBA,  described  about  the  tri- 
angle'1  DBA;  and  that  circle  is  there- 
fore given  ^  in  position :  and  the  circle 
BC  is  given  in  position,  therefore  the 
point  13  is  given^  ;  the  point  A  is  also  given;  therefore  the ^28. dat. 
straight  line  AB  is  given  *  in    position  and  magnitude. 


DB: 


because 


d6.  def. 


PROP.  XCV. 


92. 


IF  a  straight  line  be  drawn  from  a  given  point  without 
a  circle  given  in  position ;  the  rectangle  contained  by  the 
segments  betwixt  the  point  and  the  circumference  of  the 
circle  is  given. 


Let  the  straight  line  ABC  be  drawn  from  the  given  point  A 
without  the  circle  BCD  given  in  posi- 
tion, cutting  it  in  B,  C  ;  the  rectangle 
BA,  AC  is  given. 

From  the  point  A  draw  ^  AD  touch- 
ing the  circle;  therefore  AD  is  givcn^ 
in  position  and  magnitude ;  and  because 
AD  is  given,  the  square  of  AD  is  gi- 
ven <=,  which  is  equal 'i  to  the  rectangls 
rectangle  BA,  AG  is  gis'en- 


a  17. 


BA.  b94.dat- 


B  A,  AC  ;   therefore  the  ^  56.  dat. 


456  EUCLID'S 


93.  PROP.  xcvr. 

IF  a  straight  line  be  drawn  througli  a  given  point 
within  a  circle  given  in  position,  the  rectangle  contained 
by  the  segments  betwixt  the  point  and  the  circumference 
of  the  circle  is  given. 

Let  the  straight  line  BAG  be  drawn  through  the  given  point 

A  within  the  circle  BCE  given  in  position  ;    the   rectangle  BA, 

AC  is  given. 

Take  D  the  centre  of  the  circle,  join  AD, 

and  produce  it  to  the  points  E,  F ;   because 

the  points  A,  D  are  given,  the  straight  line 
a29.dat.  AD  is  given  ^  in  position  ;  and  the  circle  BEC 

is  given    in    position;    therefoi'e    the   points  B' 
b  28.  dat.  £,  p  are  given  ^ ;  and  the  point  A  is  given, 

therefore  EA,  AF  are  each  of  them  given  ^  ;       ^ 
c  35.  3.    ^"d  ^he  rectangle  EA,  AF  is  therefore  given  ;  and  it  is  equal  '^  to 

the  rectangle  BA,  AC,  which  consequently  is  given. 


M.  PROP.  XCVIL 

IF  a  straight  line  be  drawn  within  a  circle  given  in 
magnitude  cutting  off  a  segment  containing  a  given 
angle ;  if  the  angle  in  the  segment  be  bisected  by  a  straight 
line  produced  till  it  meets  the  circumference,  the  straight 
lines  which  contain  the  given  angle  shall  both  of  them 
together  have  a  given  ratio  to  the  straight  line  which 
bisects  the  angle  :  and  the  rectangle  contained  by  botli 
these  lines  together  which  contain  the  given  angle,  and 
the  part  of  the  bisecting  line  cut  off  below  the  base  of 
the  segment,  shall  be  given. 

Let  the  straight  line  BC  be  drawn  within  the  circle  ABC  gi- 
ven in  magnitude,  cutting  off  a  segment  containing  the  given 
angle  BAC,  and  let  the  angle  BAC  be  bisected  by  the  straight 
line  AD  ;  BA  together  with  AC  has  a  given  ratio  to  AD;  and 
the  rectangle  contained  by  B  A  and  AC  together,  and  the  straight 
line  ED  cut  off  from  AD  below  BC  the  base  of  the  segment,  is 
given. 

.Toiii  BD ;    and  because  BC  is  drawn  within  the  circle  ABC 


DATA. 


457 


given  in   mag;nitude  ciitting  off  the  segment  BAG,  containing 

the  given  angle  BAG  ;  BC  is  given  ^  in  magnitude  ;  by  the  same  a  91.dat. 

reason  BD  is  given  ;  therefore  '»  the  ratio  ofBC  to  BD  is  given  :  b  1.  dat. 

and  because  the  angle  BAG  is  bisected  by  AD,  as  BA  to  AC, 

so  is^BE  to  EG  ;  and,  by  permutation,  as  AB  to  BE,  so  is  AGc3.  6. 

to  GE:    wherefore  <i  as  BA  and  AC  together  to  BC,  so  is  AG  tod  12.  3. 


CE:and  because  the  angle  BAE  is  equal  to  EAG,  and  the  angle 


ACE  toe  ADB,  the  triangle  ACE  is 
equiangular  to  the  triangle'  ADB  ; 
therefore  as  AG  to  GE,  so  is  AD  to 
DB:  but  as  AC  to  GE,  so  is  BA 
together  with  AC  to  BC:  as  there- 
fore BA  and  AG  to  BC,  so  is  AD 
to  DB  ;  and,  by  permutation,  as  BA 
and  AG  to  AD,  so  is  BC  to  BD: 
and  the  ratio  of  BC  to  BD  is  given, 
therefore  the  ratio  of  BA  together  with  AC  to  AD  is  given. 

Also  the  rectangle  contained  by  BA  and  AC  together,  and 
DE  is  given. 

Because  the  triangle  BDE  is  equiangular  to  the  triangle  AGE, 
as  BD  to  DE,  so  is  AG  to  GE  ;  and  as  AC  to  GE,  so  is  BA 
and  AC  to  BC  ;  therefore  as  BA  and  AG  to  BC,  so  is  BD  to 
DE;  wherefore  the  rectangle  contained  by  BA  and  AG  together, 
and  DE,  is  equal  to  the  rectangle  GB,  BD :  but  CB,  BD  is 
given  ;  therefore  the  rectangle  contained  by  BA  and  AC  together, 
and  DE,  is  given. 


e21 


Othertvise, 


Produce  CA,  and  make  AF  equal  to  AB,  and  join  BF  ;  and 
because  the  angle  BAG  is  double  =»  of  each  of  the  angles  BFA, 
BAD,  the  angle  BFA  is  equal  to  BAD;  and  the  angle  BCA  is 
equal  to  BDA,  therefore  the  triangle  FCB  is  equiangular  to 
ABD  :  as  therefore  FC  to  GB,  so  is  AD  to  DB  ;  and,  by  per- 
mutation, as  FC,  that  is,  BA  and  AC  together,  to  AD,  so  is  CB 
to  BD  :  and  the  ratio  of  GB  to  BD  is  given,  therefore  the  ratio 
of  BA  and  AC  to  AD  is  given. 

And  because  the  angle  BFC  is  equal  to  the  angle  DAG,  that 
is,  to  the  angle  DBC,  and  the  angle  ACB  equal  to  the  angle 
ADB  ;  the  triangle  FCB  is  equiangular  to  BDE,  as  therefore 
FC  to  GB,  so  is  BD  to  DE  ;  therefore  the  rectangle  contained 
fey  FC,  that  is,  BA  and  AC  together,  and  DE  is  equal  to  the 

3  M. 


32.1. 


458  EUCLID'S 

rectangle  CB,  BD,  which  is  given,  and  therefore  the  rectangle 
contained  by  BA,  AC  together,  and  DE  is  given. 


P-  PROP.  XCVIII. 

IF  a  straight  line  be  drawn  within  a  circle  given  in 
magnitude,  cutting  oft'  a  segment  containii  g  a  given 
angle  :  if  the  angle  adjacent  to  the  angle  in  the  s?gment 
be  bisected  by  a  straight  line  produced  till  it  m.et  the 
circumference  again  and  the  base  of  the  segment ;  the 
excess  of  the  straight  lines  which  contain  the  given  an- 
gle shall  have  a  given  ratio  to  the  segment  of  the  bisect- 
ing line  which  is  within  the  circle  ;  and  the  rectangle 
,  contained  by  the  same  excess  and  the  segment  of  the 
bisecting  line  betwixt  the  base  produced  and  the  point 
where  it  again  meets  the  circumference,  shall  be  given. 


Let  the  straight  line  BC  be  drawn  within  the  circle  ABC 
given  in  magnitude  cutting  off  a  segment  containing  the  given 
angle  BAC,  and  let  tlie  angle  CAF "adjacent  to  BAG  be  bisect- 
ed by  the  straight  line  DAE  meeting  the  circumference  again 
in  D,  and  BC  the  base  of  the  segment  produced  in  E  ;  the  ex- 
cess of  BA,  AC  has  a  given  ratio  to  AD  ;  and  the  rectangle 
which  is  contained  by  the  same  excess  and  the  straight  line  ED, 
is  given. 

Join  BD,  and  through  B  draw  BG  parallel  to  DE  meeting 
AC  produced  in  G  :  and  because  BC  cuts  off  from  the  circle 
ABC  given  in  magnitude  the  seg- 
ment BAC  contaiiting  a  given  an- 
a91.dat.  S'®'  BC  is  therefore  given »  in 
magiiitude:  by  the  same  reason. 
BD  is  given,  because  the  angle 
Bad  is  equal  to  the  given  angle 
EAF  :  therefore  the  ratio  of  BC  to 
BD  is  given:  and  because  the  an- 
gle CAE  is  equal  to  EAF,  of  ^^G 
which  CAE  is  equal  to  the  alternate  angle  AGB,  and  EAF  to 
the  interior  and  opposite  angle  ABG ;  therefore  the  angle  AGB 
is  equal  to  ABG,  and  the  straight  line  AB  equal  to  AG;  so  that 


DATA.  4i9 

GC  is  the  excess  of  BA,  AC  ;  and  because  the  angle  BGC  is 
equal  to  GAE,  that  is,  to  EAF,  or  the  angle  BAD  ;  and  that 
the  angle  BCG  is  equal  to  the  opposite  interior  angle  BDA  of 
the  quadrilateral  BCAD  in  the  circle  ;  therefore  the  triangle 
BGC  is  equiangular  to  BDA  :  therefore  as  GC  to  CB,  so  is  AD 
to  DB  ;  and,  by  permutation,  as  GC  which  is  the  excess  of 
BA,  AC  to  ;\D.  so  is  CB  to  BD  :  and  the  ratio  of  CB  to  BD  is 
given  :  therefore  the  ratio  of  the  excess  of  BA,  AC  to  AD  is 
given. 

And  because  the  angle  GBC  is  equal  to  the  alternate  angle 
DEB,  and  the  angle  BCG  equal  to  BDE ;  the  triangle  BlG  is 
equiangular  to  BDE:  therefore  as  GC  to  CB,  so  is  BD  to  DE ; 
and  consequently  the  rectangle  GC,  DE  is  equal  to  the  rect- 
angle CB,  BD  which  is  given,  because  its  sides  CB,  BD  are  gi- 
ven :  therefore  the  rectangle  contained  by  the  excess  of  BA,  AC 
and  the  straight  line  DE  is  given. 


PROP.  XCIX.  9^. 

IF  from  a  given  point  in  the  diameter  of  a  circle 
given  in  position,  or  in  the  diameter  produced,  a  straight 
line  be  draA\Ti  to  any  point  in  the  circumference,  and 
from  that  point  a  straight  line  be  drawn  at  right  angles 
to  the  first,  and  from  the  point  in  which  this  meets  the 
circumference  again,  a  straight  line  be  drawn  parallel  to 
the  first ;  the  point  in  which  this  parallel  meets  the  dia- 
meter is  given ;  and  the  rectangle  contained  by  the  t^\  o 
parallels  is  given. 

In  BC  the  diameter  of  the  circle  ABC  given  in  position,  or 
in  BC  produced,  let  the  given  point  D  be  taken,  and  from  D 
let  a  straight  line  DA  be  drawn  to  any  point  A  in  the  circum- 
ference, and  let  AE  be  drawn  at  right  angles  to  DA,  and  from 
the  f>oint  E  where  it  meets  the  circumference  again  let  EF  be 
drawn  parallel  to  DA  meeting  BC  in  F;  the  point  F  is  given,  as 
also  the  rectangle  AD,  EF. 

Produce  FF  to  the  circumference  in   G,  and    join  AG  :  be- 
cause GEA  is  a  right  angle,  the  straight  line  AG  is  ^   the  dia- a  Cor,  5. 
meter  of   the  circle  ABC  ;  and    BC  is  also   a   diameter  of  U  ;*• 
therefore  the    point  H  where   they  meet   is   the  centre  of   the 
circle,  and  consequently  H  is  given  :  and  the  point  D   is  given, 
wherefore  DH  is  given  in  magnitude :  and   because  AD  is  pa- 


*^^  EUCLID'S 


b4  6. 


rallel  to  FG,   and  GH  equal  to  HA  ;  DH  is  equal  *>  to  HF,  and 
AD  equal  to  GF  :  and  DH  is  given,  therefore  HF  is  given   ia 


A  A 


inagnitude  ;  and  it  is  also  given  in  position,  and  the  point  H  is 
c  30.  dat.  given,  therefore  ^  the  point  F  is  givt-n. 

And  because    the  straight  line    KFG  is   drawn   from  a  given 

point  F  without  or  v^iihin  the  circle  ABC  given  in  position, 
d  'JS.  or  therefore  *i  the  rectangle  EF,  FG  is  given  :  and  GF  is  equal  to 
96.  dat.    ^Q^  wherefore  the  x-ectangle  AD,  EF  is  given. 

4  PROP.  c. 

IF  from  a  given  point  in  a  straight  line  given  in  posi- 
tion, a  straight  line  be  drawn  to  any  point  in  the  circum- 
ference of  a  circle  given  in  position ;  and  from  this  point 
a  straight  line  be  drawn  making  Avith  the  first  an  angle 
equal  to  the  difference  of  a  right  angle  and  the  angle 
contained  by  the  straight  line  given  in  position,  and  the 
straight  line  which  joins  the  given  point  and  the  centre 
of  the  circle ;  and  from  the  point  in  which  the  second 
line  meets  the  circumference  again,  a  third  straight  line 
be  drawn  making  with  the  second  an  angle  equal  to 
that  which  the  first  makes  with  the  second  :  the  point  in 
which  this  third  line  meets  the  straight  line  given  in  po- 
sition is  given ;  as  also  the  rectangle  contained  by  the 
first  straight  line  and  the  segment  of  the  third  betwixt 
the  circumference  and  the  straight  line  given  in  position, 
is  given. 

Let  the  straight  line  CD  be  drawn  from  the  given  point  C 
in  the  straight  line  AB  given  in  position,  to  the  circumference 
of  the  circle  DEF  given  in  position,  of  which  G  is  the  centre  ; 
join  CG,  and  from  the  point  D  let  DF  be  drawn  making  the 
angle  CDF  equal  to  the  difference  of  a  right  angle  and  the 
angle  BCG,  and   from  the  point  F  let  FE  be  drawn  making 


DATA. 


46J 


a  26. 


h  8.  i: 


the  angle  DFE  equal  to  CDF,  meeting  AB  in  H :  the  point  H 
given  ;  as  aho  the  rectangle  CD,  FH. 

Let  CD,  FH  meet  one  another  in 
the  point  K,  from  which  draw  KL 
perpendicular  to  DF;  and  let  DC  meet 
the  circumference  again  in  M,  and  let 
FH  meet  the  same  in  E,  and  join  JVIG, 
GF,  GH. 

Because  the  angles  MDF,  DFE  are 
equal  to  one  another,  the  circumfer- 
ences MF,  DE  are  equal  ^ ;  and  add- 
ing or  taking  away  the  common  part 
ME,  the  circumference  DM  is  equal 
to  EF  ;  therefore  the  straight  line  DM 
is  equal  to  the  straight  line  EF,  and 
the  angle  GMD  to  the  angle  b  GFE  ; 
and  the  angles  GMC,  GFH  are  equal 
to  one  another,  because  they  are  ei- 
ther the  same  with  the  angles  GMD, 
GFE,  or  adjacent  to  thein  :  and  be- 
cause the  angles  KDL,  LKD  are  toge- 
ther equal  <^  to  a  right  angle,  that  is, 
by  the  hypothesis,  to  the  angles  KDL, 
GCB:  the  angle  GCB,  or  GCH  is 
equal  to  the  angle  (LKD,  that  is,  to 
the  angle)  LKF  or  GKH  :  therefore  the  points  C,  K,  H,  G  are 
in  the  circumference  of  a  circle  ;  and  the  angle  GCK  is  there- 
fore equal  to  the  angle  GHF ;  and  the  angle  GMC  is  equal  to 
GFH,  and  the  straight  line  GM  to  GF  ;  therefore  ^  CG  is  equal  d  26. 1- 
to  GH,  and  CM  to  klF :  and  because  CG  is  equal  to  GH,  the 
angle  GCH  is  equal  to  GHC ;  but  the  angle  GCH  is  given  : 
therefore  GHC  is  given,  and  consequently  the  angle  CGH  is 
given  ;  and  CG  is  given  in  position,  and  the  point  G  ;  there- 
fore e  GH  is  given  in  position  ;  and  CB  is  also  given  in  position,  e  52.  dat. 
whereof  the  point  H  is  given. 

And  because  HF  is  equal  to  CM,  the  rectangle  DC,  FH  is 
equal  to  DC,  CM  :  but  DC,  CM  is  given  f,  because  the  point  C  f  95.  or 
is  given,  therefore  the  rectangle  DC,  FH  is  given.  5^-  ^^.t. 


c  32. 1. 


AC 


H  B 


FINIS. 


NOTES  ON  EUCLID'S  DATA. 


DEFINITION  II. 


This  is  made  more  explicit  than  in  the  Greek  text,  to  pre- 
vent a  mistake  which  the  author  of  the  second  demonstration  of 
the  24th  proposition  in  the  Greek  edition  has  fallen  into,  of  think- 
ing that  a  ratio  is  given  to  which  another  ratio  is  shown  to  be 
equal,  though  this  other  be  not  exhibited  in  given  magnitudes. 
See  the  Notes  on  that  proposition,  which  is  the  13th  in  this  edi- 
tion. Besides,  by  this  definition,  as  it  is  now  given,  some  pro- 
positions are  demonstrated,  which  in  the  Greek  are  not  so  well 
done  by  help  of  prop.  2. 

DEF.  IV. 

In  the  Greek  text,  def.  4.  is  thus :  "  Points,  lines,  spaces", 
"  and  angles  are  said  to  be  given  in  position  which  have  always 
"  the  same  situation  ;"  but  this  is  imperfect  and  useless,  because 
there  are  innumerable  cases  in  which  things  may  be  given  ac- 
cording to  this  definition,  and  yet  their  position  cannot  be  found ; 
for  instance,  let  the  triangle  ABC  be  given  in  position,  and  let 
it  be  proposed  to  draw  a  straight  line  BD  from  the  angle  at  B 
to  the  opposite  side  AC,  which  shall  cut  A 

off  the  angle  DBC,  which  shall  be  the 
seventh  part  of  the  angle  ABC  ;  suppose 
this  is  done,  therefore  the  straight  line 
BD  is  invariable  in  its  position,  that  is, 
has  always  the  same  situation;  for  any 
other  straight  line  drawn  from  the  point  B  on  either  side  of 
BD  cuts  off  an  angle  greater  or  lesser  than  the  'seventh  part  of 
the  angle  ABC ;  therefore,  according  to  this  definition,  tlie 
straight  line  BD  is  given  in  position,  as  also  =»  the  point  D  inj^28.  dat' 
which  it  meets  the  straight  line  AC  which  is  given  in  position. 
But  from  the  things  here  given,  neither  the  straight  line  BD 
nor  the  point  D  can  be  found  by  the  help  of  Euclid's  Elements 
only,  by  which  every  thing  in  his  Data  is  supposed  may  be  found. 


464  NOTES  ON 

This  definition  is  therefore  of  no  use.  We  have  amended  it  by 
adding-,  "and  which  are  either  actually  exhibited  or  can  be  found  ;" 
for  nothing  is  to  be  reckoned  given,  v/hich  cannot  be  found,  or  is 
not  actually  exhibited. 

The  definition  of  an  angle  given  by  position  is  taken  out  of 
th«  4th,  and  given  more  distinctly  by  itself  in  the  definition 
marked  A. 

DEF.  XI,  Xil,  XIII,  XIV,  XV. 

The  11th  and  I2th  are  omitted,  because  they  cannot  be  given 
in  English  so  as  to  have  any  tolerable  sense  ;  and,  therefore, 
wherever  the  terms  defined  occur,  the  words  which  express  their 
meaning  are  made  use  of  in  their  place. 

The  13th,   14th,   15th  are  omitted,  as  being  of  no  use. 

It  is  to  be  observed  in  general  of  the  Data  in  this  book,  that 
they  are  to  be  understood  to  be  given  geometrically,  not  always 
arithmetically,  that  is,  they  cannot  always  be  exhibited  in  num- 
bers; for  instance,  if  the  side  of  a  square  be  given,  the  ratio  of 
b.44.dat. it  to  its  diameter  i-  given''  geometrically,  but  not  in  numbers; 
c  2.  dat.  dud  the  diameter  is  given  ^  ;  but  though  the  number  of  any  equal 
parts  in  the  side  be  given,  for  example  10,  the  number  of  them 
in  the  diameter  cannot  be  given :  and  the  like  holds  in  many 
other  cases. 

PROPOSITION  I. 

In  this  it  is  shown  that  A  is  to  B,  as  C  to  D,  from  this,  that 
A  is  to  C,  as  B  to  D,  and  then  by  permutation ;  but  it  follows 
directly,  vicithout  these  two  steps,  from  7.  5. 

PROP.  II. 

The  limitation  added  at  the  end  of  this  proposition  between 
the  inverted  commas  is  quite  necessary,  because  without  it  the 
proposition  cannot  always  be  demonstrated :  for  the  author  hav- 
ing said*,  "  because  A  is  given,  a  magnitude  equal  to  it  can 
a  1.  def.  "  ^^  found  ^ ;  let  this  be  C  ;  and  because  the  ratio  of  A  to  B 
b  2.  def.  "  '*  given,  a  ratio  which  is  the  same  to  it  can  be  found  ^," 
adds,  "let  it  be  found,  and  let  it  be  the  ratio  of  C  to  A." 
Now,  from  the  second  definition  nothing  more  follows,  than 
that  some  ratio,  suppose  the  ratio  of  E  to  Z,  can  be  founds 
which  15  the  same  with  the  ratio  of  A  to  B ;  and  when  the 
author  supposes  that  the   ratio  of  C  to  A,   which  is  also   the 

•  See  Dr.  Gregory's  edition  of  tbe  Data. 


EUCLID'S  DATA.  4^ 

same  with  the  ratio  of  A  to  B,  can  be  found,  he  necessarily  sup- 
poses tliat  to  the  three  magnitudes  E,  Z,  C,  a  fourth  propor- 
tional A  may  be  found  ;  but  this  cannot  always  be  done  by 
the  Elements  of  Euclid;  from  which  it  is  plain  Euclid  must 
have  understood  the  proposition  under  the  limitation  which  is 
now  added  to  his  text.  An  example  will  make  this  clear:  let 
A  be  a  given  angle,  and  B  another  an- 
gle to  which  A  has  a  given  ratio,  for 
instance,  the  ratio  of  the  given  straight 
line  E  to  the  given  one  Z  ;  then,  hav- 
ing found  an  angle  C  equal  to  A,  how 
can  the  angle  a  be  found  to  which  C 
has  the  same  ratio  that  E  has  to  Z  ? 
certainly  no  way,  until  it  be  shown  how 
to  find  an  angle  to  which  a  given  angle 
has  a  given  ratio  which  cannot  be  done 
by  Euclid's  Elements,  nor  probably 
by  any  Geometry  known  in  his  time. 

Therefore,  in  all  the  propositions  of  this  book  which  depend  up- 
on this  second,  the  abovementioned  limitation  must  be  under- 
stood, though  it  be  not  explicitly  mentioned. 


PROP.  V. 

The  order  of  the  propositions  in  the  Greek  text,  between 
prop.  4  and  prop.  25,  is  now  changed  into  another  which  is 
more  natural,  by  placing  those  which  are  more  simple  before 
those  which  are  more  complex ;  and  by  placing  together  those 
which  are  of  the  same  kind,  some  of  which  were  mixed  among 
others  of  a  different  kind.  Thus,  prop.  12,  in  the  Greek,  is  now- 
made  the  5th,  and  those  which  were  the  22d  and  23d  are  made 
the  Uth  and  12th,  as  they  are  more  simple  than  the  proposi- 
tions concerning  magnitudes,  the  excess  of  one  of  which  above 
a  given  magnitude  has  a  given  ratio  to  the  other,  after  which 
these  two  were  placed  ;  and  the  24th  in  the  Greek  text  is,  for 
the  same  reason,  made  the  1 3th. 


PROP.  VI,  VII. 


• 


These  are  universally  true,  though,  in  the  Greek  text,  they 
are  demonstrated  by  prop.  2,  Which  has  a  limitation :  they  ar^ 
therefore  now  shown  without  it. 

3N 


m>  NOTES  ON 


PROP.  XIL 

In  the  23d  prop,  in  the  Greek  text,  which  here  is  the  ISth, 
the  words,  "  ftu  ry;  eivmi  h,"  are  wrong  translaXed  by  Claud. 
Hardy,  in  his  edition  of  Euclid's  Data,  printed  at  Paris,  anno 
1625,  which  was  the  first  edition  of  the  Greek  text  ;  and  Dr. 
Gregory  follows  him  in  transhuing  them  by  the  words,  "etsi 
non  easdem,"  as  if  the  Greek  had  been  $i  x«<  jw-ij  ?»$  xvrtif, 
as  in  prop*  9,  of  the  Greek  text.  Euclid's  mtaning  is,  that  the 
ratios  mentioned  in  the  proposition  must  not  be  the  same  ;  for, 
if  they  were,  the  proposition  would  not  be  true»  Whatever 
ratio  the  whole  has  to  the  whole,  if  the  ratios  of  the  parts  of  the 
first  to  the  parts  of  the  other  be  the  same  with  this  ratio,  one 
part  of  the  first  may  be  double,  triple,  &c.  of  the  other  part  of 
it,  or  have  any  other  ratio  to  it,  and  consequently  cannot  have  a 
given  ratio  to  it  ;  wherefore,  these  words  must  be  rendered  by 
"  non  autem  easdem,"  but  not  the  same  ratios,  as  Zambertu« 
has  translated  them  in  bis  edition. 


PROP.  XIII. 

Some  very  ignorant  editor  has  given  a  second  demonstration 
of  this  proposition  in  the  Greek  text,  which  has  been  as  igno- 
rantly  kept  in  by  Claud.  Hardv  and  Dr.  Gregory,  and  has  been 
retained  in  the  translations  of  Zambertus  and  others ;  Carolus 
Renaldinus  gives  it  only :  the  author  of  it  has  thought  that  a 
ratio  was  given  if  another  ratio  could  be  shown  to  be  the  same  to 
it,  though  this  last  ratio  be  not  found :  but  this  is  altogether  ab- 
surd, because  from  it  would  be  deduced,  that  the  ratio  of  the 
sides  of  any  two  squares  is  given,  and  the  ratio  of  the  diameters 
of  any  two  circles,  &c.  And  it  is  to  be  observed,  that  the  mo- 
derns frequently  take  given  ratios,  and  ratios  that  are  always  the 
same,  for  one  and  the  same  thing :  and  Sir  Isaac  Newton  has 
fallen  into  this  mistake  in  the  17th  lemma  of  his  Principia,  edit, 
1713,  and  in  other  places  ;  but  this  should  be  carefully  avoided, 
as  it  may  lead  into  other  errors. 


PROP.  XIV,  XV- 

Euclid,   in   this    book,    has    several   propositions    concerning 
magnitudes,  the  excess  of  one  of  which  above  a  given  magni' 


EUCLID'S  DATA.  467 

tude  has  a  given  ratio  to  the  other ;  but  he  has  given  none  con- 
cerning magnitudes  whereof  one  together  with  a  given  magni-  * 
nitude  has  a  given  ratio  to  the  other ;  though  these  last  occur 
as  frequently  in  the  solution  of  problems  as  the  first :  the  rea- 
son of  which  is,  that  the  last  may  be  all  demonstrated  by  help 
of  the  first ;  for,  if  a  magnitude,  together  with  a  given  magni- 
tude has  a  given  ratio  to  another  magnitude,  the  excess  of  this 
other  above  a  given  magnitude  shall  have  a  given  ratio  to  the 
first,  and  on  the  contrary;  as  we  have  dijmoiistrated  in  prop.  14. 
And  for  a  like  reason  prop.  15  has  been  added  to  the  Datu.  One 
example  will  make  the  thing  clear:  suppose  it  were  to  be  de- 
monstrated, that  if  a  magnitude  A  together  with  a  given  mag- 
nitude has  a  given  ratio  to  another  magnitude  B,  that  the  two 
magnitudes  A  and  B,  together  with  a  given  magnitude,  have  a 
given  ratio  to  that  other  magnitude  B ;  which  is  the  same  pro- 
position with  respect  to  the  last  kind  of  magnitudes  above-men- 
tioned, that  the  first  part  of  prop.  16,  in  this  edition,  is  in  I'e- 
spect  of  the  first  kind:  this  is  shown  thus;  from  the  hypothesis, 
and  by  the  first  part  of  prop.  14,  the  excess  of  B  above  a  given 
magnitude  has  unto  A  a  given  ratio ;  and,  therefore,  by  the  first 
part  of  prop.  17,  the  excess  of  B  above  a  given  magnitude  has 
unto  B  and  A  together  a  given  ratio ;  and  by  the  second  part  of 
prop.  14,  A  and  B  together  with  a  given  magnitude  has  unto  B 
a  given  ratio ;  which  is  the  thing  that  was  to  be  demonstrated. 
In  like  manner,  the  other  propositions  concerning  the  last  kind 
of  magnitudes  may  be  shown. 


PROP.  XVI,  XVII. 

In  the  third  part  of  prop.  10,  in  the  Greek  text,  which  is  the 
1 6th  in  this  edition,  after  the  ratio  of  EC  to  CB  has  been  shown 
to  be  given :  from  this,  by  inversion  and  conversion  the  ratio  of 
BC  to  BE  is  demonstrated  to  be  given  ;  but  without  these  two 
steps,  the  conclusion  should  have  been  made  only  by  citing  tlie 
6th  proposition.  And  in  like  manner,  in  the  first  part  of  prop. 
11,  in  the  Greek,  which  in  this  edition  is  the  17th  from  the  ratio 
of  DB  to  Be  being  given,  the  ratio  of  DC  to  DB  is  shown  to 
be  given  by  inversion  and  composition,  instead  of  citing  prop.  7, 
and  the  same  fault  occurs  in  the  second  part  of  the  same  prop. 
11. 


PROP.  XXI,  XXII. 

These  now  are  added,  as  being  wanting  to  complete  the  sub- 
ject treated  of  in  the  four  preceding  propositions. 


4&e  NOTES  ON 

PROP.  XXIII. 

This,  which  is  prop.  20,  in  the  Greek  text,  was  separated 
from  prop.  14,  15,  16,  in  that  text,  after  which  it  should  have 
been  immediately  placed,  as  being,  of  the  same  kind  ;  it  is  now 
put  into  its  proper  place  ;  but  prop.  21  in  the  Greek  is  left  out, 
as  being  the  same  with  prop.  14,  in  that  text,  which  is  here 
prop.  18. 

PPvOP.  XXIV. 

This,  which  is  prop.  13,  in  the  Greek,  is  now  put  into  its  pro- 
per place,  having  been  disjoined  from  the  three  following  it  in 
this  edition,  which  are  of  the  same  kind. 

PROP.  XXVIII. 

This,  which  in  the  Greek  text  is  prop.  25,  and  several  of  the 
following  propositions  are  there  deduced  from  def.  4,  which  is 
not  sufficient,  as  has  been  mentioned  in  the  note  on  that  defini- 
tion :  they  are  therefore  now  shown  more  explicitly. 

PROP.  XXXIV,  XXXVI. 

Each  of  these  has  a  determination,  which  is  now  added,  which 
occasions  a  change  in  their  demonstrations. 

PROP.  XXXVII,  XXXIX,  XL,  XLI. 

The  35th  and  36th  propositions  in  the  Greek  text  are  joined 
into  one,  which  makes  the  39th  in  this  edition,  because  the  same 
enunciation  and  demonstration  serves  both  :  and  for  the  same 
reason  prop.  37,  38,  in  the  Greek,  are  joined  into  one,  which 
here  is  the  40th. 

Prop.  37  is  added  to  the  Data,  as  it  frequently  occurs  in  the 
solution  of  problems  ;  and  prop.  41  is  added  to  complete  the 
rest. 

PROP.  XLII. 

This  is  prop.  39,  in  the  Greek  text,  where  the  whole  construc- 
tion of  prop.  22,  of  book  I.  of  the  Elements  is  put,  without  need, 
into  the  demonstration,  but  is  now  only  cited. 

PROP.  XLV. 

This  is  prop.  42,  in  the  Greek,  where  the  three  straight  lines 
made  use  of  in  the  construction  are  said,  but  not  shown,  to  be 
such  that  any  two  of  them  is  greater  thaa  the  third,  which,  is 
BOW  done. 


EUCLID'S  DATA.  ;4d9 

PROP.  XLVII. 

This  IS  prop.  44,  in  the  Greek  text ;  but  the  demonstration  of 
it  is  changed  into  another,  wherein  the  several  cases  of  it  are 
shown,  which,  though  .necessary,  is  not  done  in  the  Greek. 

PROP.  XLVIII. 

There  are  two  cases  in  this  proposition,  arising  from  the  two 
cases  of  the  third  part  of  prop.  47,  on  which  the  48th  depends  ; 
and  in  the  composition  these  two  cases  are  explicitly  given. 

PROP.  LII. 

The  construction  and  demonstration  of  this,  which  is  prop. 
48,  in  the  Greek,  are  made  something  shorter  than  in  that  text. 

PROP.  LIII. 

Prop.  63,  in  the  Greek  text  is  omitted,  being  only  a  case  of 
prop.  49,  in  that  text,  which  is  prop.  53,  in  this  edition. 

PROP.  LVIII. 

This  is  not  in  the  Greek  text,  but  its  demonstration  is  con- 
tained in  that  of  the  first  part  of  prop.  5  4,  in  that  text;  which 
proposition  is  concerning  figures  that  are  given  in  species:  this 
58th  is  true  of  similar  figures,  though  they  be  not  given  in  spe*' 
cies,  and  as  it  frequently  occurs,  it  was  necessary  to  add  it. 

PROP.  LIX,  LXI. 

This  is  the  54th  in  the  Greek  ;  and  the  77th  in  the  Greek, 
being  the  very  same  with  it,  is  left  out,  and  a  shorter  demon- 
stration is  given  of  prop.  61. 

PROP.  LXII. 

This,  which  is  most  frequently  useful,  is  not  in  the  Greek, 
and  is  necessary  to  prop.  87,  88,  in  this  edition,  as  also,  though 
not  mentioned,  to  prop.  86,  87,  in  the  former  editions.  Prop, 
66,  in  the  Greek  text,  is  made  a  corollary  to  it. 

PROP.  LXIV. 

This  contains  both  prop.  74,  and  73,  in  the  Greek  text  ;  the 
first  case  of  the  74th  is  a  repetition  of  prop.  55,  from  v/hicli  it 
is  separated  in  that  text  by  many  propositions  ;  and  as  there  is 
no  order  in  these  propositions,  as  they  stand  in  the  Greek,  they 
are  now  put  into  the  order  which  seemed  most  convenient  and 
natui'al. 


4ro  NOTES  ON 

The  demonstration  of  the  first  part  of  prop.  73,  in  the  Greek, 
is  grossly  vitiated.  Dr.  Gregory  says,  that  the  sentences  he  has 
inclosed  betwixt  two  stars  are  superfluous,  and  ought  to  be  can- 
celled ;  but  he  has  not  observed,  that  what  follows  them  is  ab- 
surd, being  to  prove  that  the  ratio  [see  his  figure]  of  Ai  to 
1  K  is  given,  which  by  the  hypothesis  at  the  beginning  of  the 
proposition  is  expressly  given  ;  so  that  the  whole  of  this  part 
was  to  be  altered,  which  is  done  in  this  prop.  64. 


PROP.  LXVII,  LXVIII. 

Prop.  TO,  in  the  Greek  text,  is  divided  into  these  two,  for 
the  sak.e  of  distinctness  ;  and  the  demonstration  of  the  67th  is 
rendered  shorter  than  that  of  the  first  part  of  prop.  70,  in  the 
Greek,  by  means  of  prop.  23,  of  book  6,  of  the  Elements. 


PROP.  LXX. 

This  is  prop.  62,  in  the  Greek  text ;  prop.  78,  in  that  text,  is 
only  a  particular  case  of  it,  and  is  therefore  omitted. 

Dr.  Gregory,  in  the  demonstration  of  prop.  62,  cites  the  49th 
prop.  dat.  to  prove  that  the  ratio  of  the  figure  AEB  to  the  pa- 
rallelogram AH  is  given  ;  whereas  this  was  shown  a  few  lines 
before  :  and  besides,  the  49th  prop,  is  not  applicable  to  these 
two  figures ;  because  AH  is  not  given  in  sptcies,  but  is  by  the 
step  for  which  the  citation  is  brought,  proved  to  be  given  in 
species. 


PROP.  LXXIII. 

Prop.  S3,  in  the  Greek  text,  is  neither  welj  enunciated  nor 
demonstrated.  The  73d,  which  in  this  edition  is  put  in  place  of 
it,  is  really  the  same,  as  will  appear  by  considering  [see  Dr. 
Gregory's  edition]  that  A,  B,  r,  E  in  the  Greek  text  are  four 
proportionals ;  and  that  the  proposition  is  to  show  that  a,  wliich 
has  a  given  ratio  to  E,  is  to  r,  as  B  is  to  a  straight  line  to 
which  A  has  a  given  ratio  ;  or,  by  inversibJi,  that  r  is  to  ^,  as 
a  straight  line  to  which  A  has  a  given  ratio  is  to  B  ;  that  is,  if 
the  proportionals  be  placed  in  this  order,  viz.  r,  E,  A,  B,  that 
the  first  r  is  to  A  lo  which  the  second  E  has  a  given  ratio,  as 
a  straight  line  to  which  the  third  A  has  a  given  ratio  is  to  the 
fourth  B  ;  which  is  the  enunciation  of  this  73d,  and  was  thus 
changed  that  it  might  be  made  like  to  that  of  prop.  72,  in  this 
'jcUtion,  which  is   the~82d  in  the  Greek  text  ;  and  the  demon- 


EUCLID'S  DATA.  471 

Stration  of  prop.  73  is  the  same  with  that  of  prop.  72,  only  mak- 
ing use  of  prop.  23j  instead  of  prop.  22,  of  book  5,  of  the 
Elements. 


PROP.  LXXVII. 

This  is  put  in  place  of  prop.  79,  in  the  Greek  text,  which 
is  not  a  datum,  but  a  theorem  premised  as  a  lemma  to  prop.  80 
in  that  text :  and  prop.  79  is  made  cor.  I  to  prop.  77,  in  this 
edition.  CI.  Hardy,  in  his  edition  of  the  Data,  takes  notice,  that 
in  prop.  80,  of  the  Greek  text,  the  parallel  KL  in  the  fiij;ure  of 
prop.  77,  in  this  edition,  must  meet  the  circumference,  but  does 
not  demonstrate  it,  which  is  done  here  at  the  end  of  cor.  3,  prop. 
77,  in  the  construction  for  finding  a  triangle  similar  to  ABC. 


PROP.  LXXVIII. 

The  demonstration  of  this,  which  is  prop.  80,  in  the  Greek,  is 
rendered  a  good  deal  shorter  by  help  of  prop.  77. 


PROP.  LXXIX,  LXXX,  LXXXI. 

These  are  added  to  Euclid's  Data,  as  propositions  which  are 
often  useful  in  the  solution  of  problems. 

PROP.  LXXXII. 

This,  which  is  prop.  60,  in  the  Greek  text,  is  placed  before 
the  83d  and  84th,  which,  in  the  Greek,  are  the  58th  and  59th, 
because  the  demonstration  of  these  two  m  this  edition  ar-e  dedu- 
ced from  that  of  prop.  82,  from  which  they  naturally  follow. 

PROP.  LXXXVIII,  XC. 

Dr.  Gregory,  in  his  preface  to  Euclid's  Works,  which  he 
published  at  Oxford  in  1703,  after  having  told  that  he  had  sup- 
plied the  defects  of  the  Greek  text  of  the  Data  in  innumerable 
places  from  several  manuscripts,  and  corrected  CI.  Hardy's 
translation  by  Mr.  Bernard's,  adds,  that  the  86th  theorem,  "  or 
proposition,"  seemed  to  be  remarkably  vitiated,  but  which  could 
not  be  restored  by  help  of  the  manuscripts ;  then  he  gives  three 
different  translations  of  it  in  Latin,  according  to  which  he 
thinks,  it  may  be  read  ;  the  two  first  have  no  distinct  meanings 
and  the  third,  which  he  says  is  the  best,  though  it  contains  a 


4r7'^  NOTES  ON 

tfue  proposition,  which  is  the  90th  in  this  edition,  has  no  con- 
nection in  the  least  with  the  Greek  text.  And  it  is  strange  that 
Dr.  Gregory  did  not  observe,  that,  if  prop.  86  was  changed  into 
this,  the  demonstration  of  the  86th  must  be  cancelled,  and  ano- 
ther, put  in  its  place:  but  the  truth  is,  both  the  enunciation  and 
the  demonstration  of  prop.  86  are  quite  entire  and  right,  only- 
prop.  87,  which  is  more  simple,  ought  to  have  been  placed  be- 
fore it ;  and  the  deficiency  which  the  doctor  justly  observes  to 
be  in  this  part  of  Euclid's  Data,  and  which,  no  doubt,  is  owing 
to  the  carelessness  and  ignorance  of  the  Greek  editors,  should 
have  been  supplied,  not  by  changing  prop.  86,  v/hich  is  both 
entire  and  necessary,  but  by  adding  the  two  propositionsj  which 
are  the  88th  and  90th  in  this  edition. 


PROP,  xcvni,  c. 

These  were  communicated  to  me  by  tvpo  excellent  geometers, 
the  first  of  them  by  the  Right  Honourable  the  Earl  of  Stanhope, 
and  the  other  by  Dr.  Matthew  Stewart ;  to  which  I  have  added 
the  demonstrations. 

Though  the  order  of  the  propositions  has  been  in  many  places 
changed  from  that  in  former  editions,  yet  this  will  be  of  little 
disadvantage,  as  the  ancient  geometers  never  cite  the  Data,  and 
the  moderns  very  rarely. 


As  that  part  of  the  composition  of  ^  problem  which  is  its  con- 
struction may  not  be  so  readily  deduced  from  the  analysis  by  be- 
ginners: for  their  sake  the  following  example  is  given,  in  which 
the  derivation  of  the  several  parts  of  the  construction  from  the 
analysis  is  particularly  shown,  that  they  may  be  assisted  to  do 
the  like  in  other  problems. 


PROBLEM. 

Having  given  the  magnitude  of  a  parallelogram,  the  angle  of 
which  ABC  is  given,  and  also  the  excess  of  the  square  of  its  side 
BC  above  the  square  of  tlie  side  AB ;  to  find  its  sides,  and  de- 
■scribe  it. 

The  analysis  of  this  is  the  same  with  the  demonstration  of  the 
87th  prop,  of  the  Data,  and  the  construction  lhr.t  is  given  of  the 
problem  at  the  end  of  that  proposition  is  thus  derived  from  t|^e 
analysis. 


EUCLID'S  DATA.  iti 

Let  Et^G  be  equal  to  the  given  angle  ABC,  and  because 
in  the  analysis  it  is  said  that  the  ratio  of  the  rectangle  AH, 
BC  to  the  parallelogram  AC  is  given  by  the  62d  prop.  dut. 
therefore,  from  a  point  in  FE,  the  perpendicular  EG  is  drawn 
to  FG,  as  the  ratio  of  FE  to  EG  is  the  ratio  of  the  rectangle 


F    G      L      O         H  N 


a6,  BC  to  the  parallelogram  AC  by  what  is  shown  at  the  end 
of  prop.  62.  Next,  the  magnitude  of  AC  is  exhibited  by  mak- 
ing the  rectangle  EG,  GH  equal  to  it ;  and  the  given  excess 
of  the  square  of  BC  above  the  square  of  BA,  to  v^hich  excess  the 
rectangle  CB,  BD  is  equal,  is  exhibited  by  the  rectangle  HG» 
GL :  thi,n  in  the  analysis,  the  rectangle  AB,  BC  is  said  to  be 
given,  and  this  is  equal  to  the  rectangle  FE,  GH,  because  the 
rectangle  AB.  BC  is  to  the  parallelogram  AC,  as  (FE  to  EG, 
that  is,  as  the  rectangle)  FE,  GH  to  EG,  GH  ;  and  the  paral» 
lelogram  AC  is  equal  to  the  rectangle  EG,  GH,  therefore  thft 
rectangle  AB,  BC,  is  equal  to  FE,  CiH:  and  consequently  the 
ratio  of  the  I'ectangle  CB,  BD,  that  is,  of  the  rectangle  HG, 
GL,  to  AB,  BC,  that  is,  of  the  straight  line  DB  to  BA,  is 
the  same  with  the  ratio  (of  the  rectangle  GL,  GH  to  FE,  GH, 
that  is)  of  the  straight  line  GL  to  FE,  which  ratio  of  DB  to 
BA  is  the  next  thing  said  to  be  given  in  the  analysis ;  from 
this  it  is  plain  that  the  square  of  FE  is  to  the  square  of  GL,  as 
the  square  of  BA,  which  is  equal  to  the  rectangle  BC,  CD,  is 
to  the  square  of  BD :  the  ratio  of  which  spaces  is  the  next 
thing  said  to  be  given  :  and  from  this  it  follows  that  four  times 
the  square  of  FE  is  to  the  square  of  GL,  as  four  times  the  rect- 
angle BC,  CD  is  to  the  square  of  BD  ;  and,  by  composition, 
four  times  the  square  of  FE  together  with  the  square  of  GL, 
is  to  the  square  of  GL,  as  four  times  the  i'ectangle  BC,  CD, 
together  with  the  square  of  BD,  is  to  the  square  of  BD,  that 
is  (8.  6,)  as  the  square  of  the  straight  lines  BC,  CD  taken  to- 
gether is  to  the  square  of  BD,  which  ratio  is  the  next  thing 
said  to  be  given  in  the  analysis:  and  because  four  times  the 
square  of  FE  and  the  square  of  GL  are  to  be  added  together  J 
therefore  in  the  perpendicular  EG  ther&  is  taken  KG  equal  to 

30 


NOTES  ON 

FE,  and  iMG  equal  to  the  double  of  it,  because  thereby  the 
squares  of  MG,  GL,  that  is,  joining  ML,  the  square  of  ML 
is  equal  to  four  limes  the  square  of  FE  and  to  the  square  of 
GL :  and  because  the  square  of  ML  is  to  the  square  of  GL,  as 
the  square  of  the  straight  line  made  up  of  BC  and  CD  is  to  the 
square  of  BD,  therefore  (22.  6.)  ML  is  to  LG,  as  BC  together 
•with  CD  is  to  BD ;  and,  by  composition,  ML  and  LG  together, 
tliat  is,  producing  GL  to  N,  so  that  ML  be  equal  to  LN,  the 
straight  line  NG  is  to  GL,  as  twice  BC  is  to  BD  ;  and  by  taking 
GO  equal  to  the  half  of  NG,  GO  is  to  GL,  as  BC  to  BD,  the 
ratio  of  which  is  said  to  be  given  in  the  analysis :  and  from  this 
it  follows,  that  the  rectangle  HG,  GO  is  to  HG,  GL,  as  the 
square  of  BC  is  to  the  rectangle  CB,  BD,  which  is  equal  to  the 
lectangle  HG,  GL ;  and  therefore  the  square  of  BC  is  equal  to 
the  rectangle  HG,  GO  ;  and  BC  is  consequently  found  by  taking 
a  mean  proportional  betwixt  HG  and  GO,  as  is  said  in  the  con- 
struction :  and  because  it  was  shown  that  GO  is  to  GL,  as  BC 
to  BD,  and  that  now  the  three  first  are  found,  the  fourth  BD  is 
found  by  12.  6.  It  was  likewise  shown  that  LG  is  to  FE,  or 
GK,  as  DB  to  BA,  and  the  three  first  are  now  found,  arid  there- 
by the  fourth  BA.  Make  the  afigle  ABC  equal  to  EFG,  and 
complete  the  parallelogram  of  which  the  sides  are  AB,  BC,  and 
the  construction  is  finished ;  the  rest  of  the  composition  con- 
lains  the  demonstration. 


AS  the  propositions  from  the  13th  to  the  28th  may  be  thought 
by  beginners  to  be  less  useful  ihan  the  rest,  because  they  can- 
not so  readily  see  how  they  are  to  be  made  use  of  in  the  solution 
of  problems  ;  on  this  account  the  two  following  problems  are 
added,  to  show  that  they  are  equally  useful  with  the  other  pro- 
])ositions,  and  from  which  it  may  be  easily  judged  that  many 
other  problems  depend  upon  these  propositions. 


PROBLEM  I. 

To  find  three  straight  lines  such,  that  the  ratio  of  the 
first  to  the  second  is  given ;  and  if  a  given  straight  line 
be  taken  from  the  second,  the  ratio  of  the  remainder  to 
the  third  is  given ;  also  the  rectangle  contained  by  the 
first  and  third  is  given. 


EUCLID'S  DMA.  475 

Let  AB  be  the  first  straight  line,  CD  the  second,  and  EF  the 
third  ;  and  because  the  ratio  of  AB  to  CD  is  given,  and  that 
if  a  given  straight  line  be  taken  from  CD,  the  ratio  of  the  re- 
mainder to  EF  is  given  ;  therefore  ^  the  excess  of  flie  first  ABa24.  dar. 
above  a  given  straight  line  has  a  given  ratio  to  the  third  EF  : 
let  BH  be  that  given  straight  line  ;  therefore  AH,  the  excess 
of  AB  above  it,  has  a  given  ratio  to  EF :  and 

consequently^  the  rectangle  BA,  AH,  has  a     A  H     B    bl.6. 

given  ratio  to  the  rectangle  AB,  EF,   which 
last  rectangle   is   given   by    the    hypothesis  ; 

therefore  ^  the  rectangle  BA,   AH  is   given,     ^  ^     U    £.  2.  dat. 

and  BH  the  excess  of  its  sides  is  given  ;  where- 


A 

H 

1 

B 

C 

1 

G 

i 

D 

E 

1 
F 

K 

NM  L 

-1-1-1- 

O 

fore  the  sides   AB,  AH  are   given  d;  and  be-     -p  y,  d85.dat. 

cause  the  ratios  of  AB  to  CD,  and  of  AH  to 
EF  are  given,  CD  and  EF  are  <=  given. 

The  Composition* 

Let  the  given  ratio  of  KL  to  KM  be  that  which  AB  is  requir- 
ed to  have  to  CD  ;  and  let  DG  be  tlie  given  straight  line  wliich 
is  to  be  taken  from  CD,  and  let  the  given  ratio  of  KM  to  KN  be 
that  which  the  remainder  must  have  to  EF  ;  also  let  the  given 
rectangle  NK,  KO  be  that  to  which  the  rectangle  AB,  EF  is  re- 
quired to  be  equal:  find  the  given  straight  line  BH  which  is  to 
be  taken  from  AB,  which  is  done,  as  plainly  appears  from  prop. 
24,  dat.  by  making  as  KM  to  KL,  so  GD  to  HB.  To  the  given 
straight  line  BH  apply  e  a  rectangle  equal  to  LK,  KO  exceeding  e  29.  C>, 
by  a  square,  and  let  BA,  AH  be  its  sides  :  then  is  AB  the  first 
of  the  straight  lines  required  to  be  found,  and  by  making  as  LK 
to  KM,  so  AB  to  DC,  DC  will  be  the  second  :  and  lastly,  make 
as  KM  to  KN,  so  CG  to  EF,  and  EF  is  the  third. 

For  as  AB  to  CD.  so  is  HB  to  GD,  each  of  these  ratios  being 
the  same  with  the  ratio  of  LK  to  KM  ;  therefore  ^  AH  is  to  CG,  f  19. 5, 
as  (AB  to  CD,  that  is,  as)  LK  to  KM  ;  and  as  CG  to  EF,  so  is 
KM  to  KN  ;  wherefore,  ex  squali,  as  AH  to  EF,  so  is  LK  to 
KN :  and  as  the  rectangle  BA,  AH  to  the  rectangle  BA,  EF,  so. 
is  g  the  rectangle  LK,  KO  to  the  rectangle  KN,  KO  :  and  by  the  g  i.  6. 
construction,  the  rectangle  BA,  AH  is  equal  to  LK,  KO  :  there- 
fore ^  the  rectangle  AB,  EF  is  equal  to  the  given  rectangle  NK,  hl4.  ^. 
KO  :  and  AB  has  to  CD  the  given  ratio  of  KL  to  KM  ;  and  iVoin 
CD  the  given  straight  line  GD  being  taken,  the  remainder  CG 
has  to  EF  the  given  ratio  of  KM  to  KN.    Q.  E.  D. 


476. 


NOTES  ON 


a  2*.dat 


b  44.  dat 


c  32.  dat, 
d47.  i 

e  34  dat 


£28.  dat 
g33  dat 

h  29  dat 
i  «.  dat. 


PROB.  II. 

TO  find  three  straight  lines  such,  that  the  ratio  of  the 
first  to  the  second  is  given ;  and  if  a  given  straight  Hne 
be  taken  from  the  second,  the  ratio  of  the  remainder  to 
the  third  is  given ;  also  tlie  sum  of  the  squares  of  the 
first  and  third  is  given. 

Let  AB  be  the  first  straight  Hne,  BC  the  second,  and  BD  the 
third  :  and  because  the  ratio  of  AB  to  BC  is  given,  and  that  if 
a  given  straight  line  be  taken  from  BC,  the  ratio  of  the  remain- 

,  der  to  BD  is  given ;  therefore  ^  the  excess  of  the  first  AB  above 
a  given  straight  line,  has  a  given  ratio  to  the  third  BD  :  let  AE 
be  that  given  straight  line,  therefore  the  i*emainder  EB  has  a 
given  ratio  to  BD  :  let  BD  be  placed  at  right  angles  to  EB  and 
join  DE  ;  then  the  triangle  EBD  is  ^  given  in  species;  where- 
fore the  angle  BE^D  is  given  :  let  AE,  which  is  given  in  magni- 
tude, be   given  also  in  position,   as  also  the  point  E,  and  the 

,  straight  line  ED  will  be  given  ^  in  position  :  join  AD,  and  be- 
cause the  sum  of  the  squares  of  AB,  BD,  that  h^,  the  square 
of  AD  is  givtn,  therefore  the  straight  line  AD  is  given  in  mag- 

.  nitude  ;  and  it  is  also  i^iven  e  in  position,  because  from  the  given 
point  A  it  is  drawn  to  the  straight  line  ED  given  in  position  : 
therefore  the  point  D,  in  which  the  two  straight  lines   AD,  ED 

•  given  in  position  cut  one  another,  is  given  f;   and  the   straight 

,  line  DB  which  is  at  right  angles  to  AB  is  given  s  in  position,  and 
AB  is  given  in  position,  therefore  f  the  point  B  is  given :  and  the 

.  points  A,  D  are  given,  wherefore  ^  the  straight  lines  AB,  BD  are 
given:  and  the  ratio  of  AB  to  BC  is  given,  and  therefore*  BC 
is  criven. 


The  Corr.fiosUion, 

Let  the  given  ratio  of  FG  to  GB  be  that  which  AB  is  requir- 
ed to  have  to  BC,  and  let  HK  be  the  given  straight  line  which  is 
to  be  taken  from  BC,  -dud  let  the  ratio  which  the  remainder  is 

L 
D 


B    N    ISl 


req'.iircd  to  have  to  BD  be   the  given   ratio  of  HG  to   GL,  and 
plaqe  GL  at  right  angles  to  FH,  and  join  LF,  LH :  next,  as  HG 


EUCLID'S  DATA.  477 

* 

is  to  GF,  so  make  HK  to  AE ;  produce  AE  to  N,  so  that  AN 
be  the  straight  line  to  the  square  of  which  the  sum  of  the  squares 
of  AB,  BD  is  required  to  be  equal ;  and  make  the  angle  NED 
equal  to  the  angle  GFL ;  and  from  the  centre  A  at  the  distance 
AN  describe  a  circle,  and  let  its  circumference  meet  ED  in  D, 
and  draw  DB  perpendicular  to  AN,  and  DM,  making  the  angle 
BDM  equal  to  the  angle  GLH.  Lastly,  produce  BM  to  C,  so 
that  MC  be  equal  to  HK.  then  is  AB  the  first,  BC  the  second, 
and  BD  the  third  of  the  straight  lines  that  were  to  be  found. 

For  the  triangles  EBD,  FGL,  as  also  DBM,  LGH  being  eqi- 
angular,  as  EB  to  BD,  so  is  FG  to  GL  ;  and  as  DB  to  BM,  so 
is  LG  to  GH ;  therefore,  ex  aequali,  as  EB  to  BM,  so  is  (FG  to 
GH,  and  so  is)  AE  to  HK  or  MC  ;  wherefore  \  AB  is  to  BC,  k  12.  5> 
as  AE  to  HK.  that  is,  as  FG  to  GH,  that  is,  in  the  given  ratio ; 
and  from  the  straight  line  BC  taking  MC,  which  is  equal  to  the 
given  straight  line  HK,  the  remainder  BM  has  to  BD  the  given 
ratio  of  HG  to  GL ;  and  the  sum  of  the  squares  of  AB,  BD  is 
equal  d  to  the  square  of  AD  or  AN,  which  is  the  given  space,  d  47. 1. 
Q.  E.  D. 

I  believe  it  would  be  in  vain  to  try  to  deduce  the  preceding 
construction  from  an  algebraical  solution  of  the  problem. 


FINIS. 


I 


TBS 


ELEMENTS 


OF 


PLANE  AND  SPHERICAL 


TRIGONOMETRY. 


J  ■  ■■!■' 


PHILADELPHIA  J 

PRINTED    FOR,    AND    PUBLISHED  BT,    MATHEW   CARET, 
NO.  122,   MARKET-STREET, 

18Q6, 


PLANE  TRIGONOMETRY. 


LEMMA  I.  FIG.   1. 


IjET  ABC  be  a  rectilineal  angle,  if  about  the  point  li  as  a  cen- 
tre, and  with  any  distance  BA,  a  circle  be  described,  meeting 
BA,  BC,  the  straight  lines  including  the  angle  ABC  in  A,  C; 
the  angle  ABC  will  be  to  four  right  angles,  as  the  arch  AC  fi 
the  whole  circumference. 

Produce  AB  till  it  meet  the  circle  again  in  F,  and  through  B 
"draw  DE  perpendicular  to  AB,  meeting  the  circle  in  D,  E. 

By  33.  6.  Elem.  the  angle  ABC  is  to  a  right  angle  ABD,  as 
the  arch  AC  to  the  arch  AD;  and  quadrupling  the  consequents, 
the  angle  ABC  will  be  to  four  right  angles,  as  the  arch  AC  td 
four  time3  the  arch  AD,  or  to  the  whole  circumference. 

LEMMA  II.  FIG.  2. 

LET  ABC  be  a  plane  rectilineal  angle  as  before  ;  alxiut  B  jls  a 
centre  with  any  two  distances  BD,  BA,  let  two  circles  be  describ- 
ed meeting  BA,  BC,  in  D,  E,  A,  C  ;  the  arch  AC  will  be  to  thr. 
whole  circumference  of  which  it  is  an  arch,  as  the  arch  DE  is 
lo  the  whole  circumference  of  which  it  is  an  arch. 

By  Lemma  1.  the  arch  AC  is  to  the  whole  circumference  of 
which  it  is  an  arch,  as  the  angle  ABC  is  to  four  right  angles  ; 
and  by  the  same  Lemma  1.  the  arch  DE  is  to  the  whole  circum- 
ference of  which  it  is  an  arch,  as  the  angle  ABC  is  to  four  right 
angles ;  therefore  the  arch  AC  is  to  the  whole  circumference  of 
which  it  is  an  arch,  as  the  arch  DE  to  the  whole  circumference 
of  which  it  is  an  arch. 

DEFINITIONS.    FIG.  5> 

I. 

LET  ABC  be  a  plane  rectiHrteal  angle  ;  if  about  B  as  a  centre^, 
with  BA  any  distance,  a  circle  ACF  be  described  meeting  BA< 
BC  in  A,  C  ;  the  arch  AC  is  called  the  measure  of  the  angle 
ABC. 

II. 

The  rircumlerrnce  of  a  circle  is  snpp'o'^cd  to  be  divided  into 


PLANE  TRIGONOMETRY. 

360  equal  parts  called  degrees,  and  each  degree  into  60  equal 
parts  called  minutes,  and  each  minute  into  60  equal  parts  cal- 
led seconds,  &c.  And  as  many  degrees,  minutes,  seconds,  &c. 
as  are  contained  in  any  arch,  of  so  many  degrees,  minutes,  se- 
conds, &c.  is  the  angle,  of  which  that  arch  is  the  measure, 
said  to  be. 
Cor.  Whatever  be  the  radius  of  the  circle  of  which  the  measure 
of  a  given  angle  is  an  arch,  that  arch  will  contain  the  same 
mmiber  of  degrees,  minutes,  seconds,  8cc.  as  is  manifest  from 
Lemma  2. 

in. 

Let  AB  be  produced  till  it  meet  the  circle  again  in  E,  the  angl^ 
CBF,  which,  together  with  ABC,  is  equal  to  two  right  angles, 
is  culled  the  Suplilement  of  the  angle  ABC. 

IV. 

A  straight  line  CD  drawn  through  C,  one  of  the  extremities  of 
the  arch  AC  perpendicular  upon  the  diameter  passing  through 
the  other  extremity  A,  is  called  the  Sine  of  the  arch  AC,  oj- 
of  the  angle  ABC,  of  which  it  is  the  measure. 

CoK.  The  Sine  of  a  quadrant,  or  of  a  right  angle,  is  equal  to  the 
radius. 

V. 

The  segment  DA  of  the  diameter  passing  through  A,  one  ex- 
tremity of  the  arch  AC  between  the  sine  CD,  and  that  ex- 
tremity, is  called  the  Versed  Si?ie  of  the  arch  AC,  or  angle 
ABC. 

VI. 

A  straight  line  AE  touching  the  circle  at  A,  one  extremity  of 
the  arch  AC,  and  meeting  the  diameter  BC  passing  through 
the  other  extremity  C  in  E,  is  called  the  Tangent  of  the  arch 
AC,  or  of  the  angle  ABC. 

vn. 

The  straight  line  BE  between  the  centre  and  the  extremity  of 
the  tangent  AE,  is  called  the  Secant  oS.  the  arch  AC,  or  an- 
gle ABC. 

Cor.  to  def.  4,  6,  7.  the  sine,  tangent,  and  secant  of  any  angle 
ABC,  are  likewise  the  sine,  tangent,  and  secant  of  its  supple- 
ment CBF. 

It  is  manifest  from  def.  4.  that  CD  is  the  sine  of  the  angle  CBF. 
Let  CB  be  produced  till  it  meet  the  circle  again  in  G ;  and  it 
is  manifest  that  AE  is  the  tangent,  and  BE  the  secant,  of  the. 
angle  ABG  or  EBF,  from  def.  6,  7. 


PLANE  TRIGONOMETRY.  48; 

C©R.  to  def.  4,  5,  6,  7.  The  sine,  versed  sine,  tangent,  and  se- 
cant, of  any  arch  which  is  the  measure  of  any  given  angle 
ABC,  is  to  the  sine,  versed  sine,  tangent,  and  secant,  of  any 
other  arch  which  is  the  meysure  of  the  same  angle,  as  the 
radius  of  the  first  is  to  the  radius  of  the  second. 

Let  AC,  MN  be  measures  of  the  angle  ABC,  according  to 
def.  1.  CD  the  sine,  DA  the  versed  sine,  AE  the  tangent 
and  BE  the  secant  of  the  arch  AC,  according  to  def.  4,  5,  6, 
7.  and  NO  the  sine,  OM  the  versed  sine,  MP  the  tangv  ut, 
and  BP  the  secant  of  the  arch  MN,  according  to  the  same 
definitions.  Since  CD,  NO,  AE,  MP  are  parallel,  CD  is  to 
NO  as  the  radius  CB  to  the  radius  NB,  and  AE  to  MP  as 
AB  to  BM,  and  BC  or  BA  to  BD  as  BN  or  BM  to  BO ;  and, 
by  conversion,  DA  to  MO  as  AB  to  MB.  Hence  the  corol- 
lary is  manifest ;  therefore,  if  the  radius  be  supposed  to  be 
divided  into  any  given  numbvrof  equal  parts,  the  sine,  versed 
sine,  tangent,  and  secant  of  any  given  angle,  will  each  contain 
a  given  number  of  these  parts;  and,  by  trigonometrical  tables, 
the  length  of  the  sine,  versed  sine,  tangent,  and  secant  of  any 
angle  may  be  found  in  parts  of  which  the  radius  contains  a 
given  number;  and,  vice  versa,  a  number  exprcasing  the  length 
of  the  sine,  versed  sine,  tangent,  and  secant  being  given,  the 
angle  of  which  it  is  the  sine^  versed  sine,  tangent,  and  secant 
may  be  found. 

VilL  Fig.  S 

The   difference   of  an  angle   from  a  right  angle  is  called  the 
coml}lement  of  that  angle.     Thus,  if  BH  be  drawn   perpendi- 
cular to  AB,  the  angle  CBH  will  be  the  complement  of  the     . 
angle  ABC,  or  of  CBF. 

IX. 

Let  HK  be  the  tangent,  CL  or  DB,  which  is  equal  to  it,  the  sine, 
and  BK  the  secant  of  CBH,  the  complement  of  ABC,  accord- 
ing to  def.  4,  6,  7.  HK  is  called  the  co-tangent,  BD  the  co-sine, 
and  BK  the  co-secant  of  the  angle  ABC. 

CoR.  1.  The  radius  is  a  mean  proportional  between  the  tangent, 
and  co-tangent. 

For,  since  HK,  BA  are  parallel,  the  angles  HKB,  ABC  will  be 
equal,  and  the  angles  KHB,  BAE  are  right ;  therefore  the  tri- 
angles BAE,  KHB  are  similar,  and  therefore  AE  is  to  AB,  as 
BH  or  BA  to  HK. 

CoR.  2.  The  radius  is  a  mean  proportional  between  the  co-sine 
and  secant  of  any  angle  ABC. 

S^ncf.  CD,  AE  are  parallel,  BD  is  to  BC  or  BA,  as  BA  to  BE. 


*ti-i  PLANE  TRIGONOMETRY. 


PRO?.  I.  FIG.  5. 


IN  a  riglit  anf^lcd  plane  triangle,  if  the  liypotheniise 
be  made  radius,  the  sides  become  the  sines  oi'  the  angles 
opposite  to  them ;  and  if  either  side  be  made  radius,  the 
remaining  side  is  the  tangent  of  the  angle  opposite  to  it^ 
and  the  hypothenuse  the  secant  of  the  same  angle. 

Let  ABC  be  a  right  angled  triangle ;  if  the  hypothenuse  BC 
be  made  radius,  either  of  the  sides  AC  will  be  the  sine  of  the 
angle  ABC  opposite  to  it;  and  if  either  side  BA  be  made  ra- 
dius, the  other  side  AC  will  be  the  tangent  of  the  angle  ABC 
opposite  to  it,  and  the  hypothenuse  BC  the  secant  of  the  same 
angle. 

About  B  as  a  centre,  with  BC,  BA  for  distances,  let  two  circles 
CD,  EA  be  described,  meeting  BA,  BC  in  D,  E :  since  CAB  is 
a  right  angle,  BC  being  radius,  AC  is  the  sine  of  the  angle 
ABC  by  def.  4.  and  BA  being  radius,  AC  is  the  tangent,  and 
BC  the  secant  of  the  angle  ABC,  by  def.  6,  7. 

Cor.  J.  Of  the  hypothenuse  u  side  and  an  angle  of  a  right 
angled  triangle,  any  two  being  given,  the  third  is  alsp  given. 

Cor.  Of  the  two  sides  and  an  angle  of  a  right  angled  triangle, 
^ny  two  being  given,  the  third  is  aho  given. 


PROP.  n.  FIG.  6,  7. 

THE  sides  of  a  plane  triangle  arc  to  one  anotiier  as  the- 
bines  of  the  angles  opposite  to  them. 

In  right  angled  triangles,  thjs  prop,  is  manifest  from  prop.  1. 
for  if  the  hypothenuse  be  made  radius,  the  sides  are  the  sines  of 
the  angles  opposite  to  them,  and  the  radius  is  the  sine  of  a  right 
angle  (cor.  to  def.  4.))  which  is  opposite  to  the  hypothenuse. 

In  any  oblique  angled  triangle  ABC,  any  two  sides  AB,  AC 
will  be  to  one  another  as  the  sines  of  the  angles  ACB,  ABC 
Avhich  are  opposite  to  thenri. 

From  C,  B  draw  CE,  BD  perpendicular  upon  the  opposite 
sides  AB,  AC  produced,  if  need  be.  Since  CEB,  CDB  are 
right  angles,  BC  being  radius,  CE  is  the  sine  of  the  angle  CBA, 
and  BD  the  sine  of  the  angle  ACB;  but  the  two  triangles  CAE, 
DAB  have  each  a  right  angle  at  D  and  E  ;  and  likewise  the 
cqmmou  angle  CAB;  therefore  they  are  liiaiilar,  i.nd   tynsc- 


PLANE  TRIGONOMETRY,.  48^ 

.^uently,  CA  is  to  AB,  as  CE  to  DB ;  that  is,  the  side^  are  as 
tt»e  sines  of  the  angles  opposite  to  them. 

Cor.  Hence  of  two  sides,  and  two  angles  opposite  to  them, 
in  a  plane  triangle,  any  three  being  given,  the  fourth  is  also 
§;iven. 


PROP.  III.  FIG.  8. 

In  a  plane  triangle,  the  sum  of  any  t^\o  sides  is  to 
tlieir  difference,  as  the  tangent  of  half  the  sum  of  tlu; 
angles  at  the  base,  to  the  tangent  of  half  their  differ.-, 
ence. 

Let  ABC  be  a  plane  triangle,  the  sum  of  any  two  sides,  AB, 
AC  will  be  to  their  diflercnce  as  the  tangent  of  half  the  sum  of 
the  angles  at  the  base  ABC,  ACB  to  the  tangent  of  half  their 
difference. 

About  A  as  a  centre,  with  AB  the  greater  side  for  a  distance, 
let  a  circle  be  described,  meeting  AC  produced  in  E,  F,  and  BC 
in  D ;  join  DA,  EB,  FB ;  and  draw  FG  parallel  to  BC,  meeting 
EB  in  G. 

The  angle  EAB  (32.  1.)  is  equal  to  the  sum  of  the  angles  at 
the  base,  and  the  angle  EFB  at  ihe  circumference  is  equal  to  the 
half  of  EAB  at  the  centre  (20.  3.);  therefore  EFB  is  half  the 
sum  of  the  angles  at  the  base;  but  the  angle  ACB  (32.  1.)  is 
equal  to  the  angles  CAD  and  ADC,  or  ABC  together ;  therefore 
Fad  is  the  difference  of  the  angles  at  the  base,  and  FBD  at  the 
circumference,  or  BFG,  on  account  of  the  parallels  FG,  BD,  is 
the  half  of  that  difference ;  but  since  the  angle  EBF  in  a  semi- 
circle is  a  right  angle  (I.  of  this)  FB  being  radius,  BE,  BG,  are 
the  tangents  of  the  angles  EFB,  BFG ;  but  it  is  manifest  that 
EC  is  the  sum  of  the  sides  BA  AC,  and  CF  their  difference  ; 
and  since  BC  FG  are  parallel  (2*.  6.)  EC  is  to  CF,  as  EB  to  BG  ; 
that  is,  the  sum  of  the  sides  is  to  their  difference,  as  t!ie  tangent 
of  half  the  sum  of  the  angles  at  the  base  to  the  tangent  of  hal! 
their  difference. 


PROP.  IV.   FIG.  18. 

In  any  plane  triangle  BAG,  ^\hose  Uvo  sides  aie  BA, 
AC,  and  base  BC,  the  less  of  the  two  sides  which  let  be 
BA,  is  to  the  greater  AC  as  the  radius  is  to  the  tangent 
pf  an  r.n^Ic,  and  the  radius  is  to  the  tangeixt  of  tlie  ex- 


486  PLANE  TRIGONOMETRY. 

cess  of  this  angle  above  half  a  right  angle  as  the  tangent 
of  half  the  supi  of  the  angles  B  and  C  at  the  base,  is  to 
the  tansrent  of  half  their  difference. 

At  the  point  A,  draw  the  straight  line  EAD  perpendicular  to 
BA;  make  AE,  AF,  each  equal  to  AB,  and   AD  to  AC;  join 

BE,  BF,  BD,  and  from  D  draw  DG  perpendicular  upon  BF. 
And  because  BA  is  at  right  angles  to  EF,  and  EA,  AB,  AF  are 
equal,  each  of  the  angles  EBA  ABF  is  half  a  right  angle,  and 
the  whole  EBF  is  a  right  angle;  also  (4.  1.  El.)  EB  is  equal  to 

BF.  And  since  EBF,  FGD  are  right  angles,  EB  is  parallel  to 
CD,  and  the  triangles  EBF,  FGD  are  similar;  therefore  EB  is 
to  BF  as  DG  to  GF,  and  EB  being  equal  to  BF,  FG  must  be 
equal  to  GD.  And  because  BAD  is  a  right  angle,  BA  the  less 
side  is  to  AD  or  AC  the  greater,  as  the  radius  is  to  the  tangent 
of  the  angle  ABD  ;  and  because  BGD  is  a  right  a'ngle,  EG  is 
to  GD  or  GF  as  the  radius  is  to  the  tangent  of  GBD,  which  is 
the  excess  of  the  angle  ABD  above  ABF  half  a  right  angle.  But 
because. EB  is  parallel  to  GD,  BG  is  to  C^F  as  ED  is  to  DF,  that 
is,  since  ED  is  the  sum  of  the  sides  BA,  AC  and  FD  *heir  dif- 
ference (3.  of  this),  as  the  tangent  of  half  the  sum  ot  the  angles 
B,  C,  at  the  base  to  the  tangent  of  half  their  difference.  There- 
fore, in  any  plane  triangle?  Stc     Q.  E.  D. 


PROP.  V.    FIG.  9,  10. 

IN  any  plane  triangle,  twice  the  rectangle  contained  by 
any  two  sides  is  to  the  difference  of  the  surn  of  the  squares 
of  these  tw^o  sides,  and  the  sqnarc  of  the  base  as  the  ra- 
dius is  to  the  co-sine  of  the  angle  included  by  the  two 
sides. 

Let  ABC  be  a  plane  triangle,  twice  the  rectangle  ABC  con- 
to.ined  by  any  two  sides  BA,  BC  is  to  the  differencs  of  the  sum 
of  the  squares  of  Bx\,  BC,  and  the  square  of  the  base  AC,  as 
the  radius  to  the  co-sine  of  the  angle  ABC. 

From  A,  draw  AD  perpendicular  upon  the  opposite  side  BC, 
then  (by  12.  and  13.  2.  El.)  the  difference  of  the  sun\  of  the 
squares  of  AB,  BC,  and  the  square  of  the  base  AC,  is  equal  to 
twice  the  rectangle  CBD;  but  twice  the  rectangle  CBA  is  to 
twice  the  rectangle  CBD;  that  is,  to  the  difference  of  the  sura 


PLANE  TRIGONOMETRY.  t%7 

of  the  squares  of  AB,  BC,  and  the  square  of  AC,  (1.  6.)  as  AB 
to  BD  ;  that  is,  by  prop.  I.  as  radius  to  the  sine  of  BAD,  which 
is  the  complement  of  the  avigle  ABC,  that  is,  as  radius  to  the 
co-sine  of  ABC. 

FROP.  VI.  FIG.  11. 

IN  any  plane  triangle  ABC,  whose  t^^o  sides  arc  AB, 
AC,  and  base  BC,  the  rectangle  contained  by  half  the  peri- 
meter, and  the  excess  of  it  abo^e  the  base  BC,  is  to  the 
rectangle  contained  by  the  sti-aight  lines  by  AA'hich  the  hall' 
of  the  perimeter  exceeds  the  other  two  sides  AB,  AC, 
as  the  square  of  the  radius  is  to  the  square  of  the  tangent 
of  half  the  angle  BAC  opposite  to  the  base. 

Let  the  angles  BAC,  ABC  be  bisected  by  the  straight  lines 
AG,  BG  ;  and  producing  the  side  AB.  let  the  exterior  angle 
CBH  be  bisected  by  the  straight  line  BK,  meeting  AG  in  K; 
and  from  the  points  G,  K,  let  there  be  drawn  perpendicular 
upon  the  sides  the  straight  lines  GD,  GE,  GF,  KH,  KL,  KM. 
Since  therefore  (4.  4.)  G  is  the  centre  of  the  circle  inscribed  in 
the  triangle  ABC,  GD,  GF,  GE  will  be  equal,  and  AD  will  be 
equal  to  AE,  BD  to  BF,  and  CE  to  CF.  In  like  manner  KH, 
KL,  KM  will  be  equal,  and  BH  will  be  equal  to  BM,  and  AH 
to  AL,  because  the  angles  HBM,  HAL  are  bisected  by  the 
straight  lines  BK,  KA  :  and  because  in  the  triangles  KCL, 
KCM,  the  sides  LK,  KM  are  equal,  KC  is  common  and  KLC, 
KMC  are  right  angles,  CL  will  be  equal  lo  CM :  since  there- 
fore BM  is  equal  to  BH,  and  CM  to  CL ;  BC  will  be  equal  to 
BH  and  CL  together;  and,  adding  AB  and  AC  together,  AB, 
AC,  and  BC  will  together  be  equal  to  AH  and  AL  together  : 
but  AH,  AL  are  equal :  wherefore  each  of  them  is  equal  to  half 
the  perimeter  of  the  triangle  ABC :  but  since  AD,  AE  arc 
equal,  and  BD,  BF,  and  also  CE,  CF,  AB  together  with  FC, 
will  be  equal  to  half  the  perimeter  of  the  triangle  to  which  AH 
or  AL  was  shewn  to  be  equal ;  taking  away  therefore  the  com- 
mon AB,  the  I'emainder  FC  will  be  equal  to  the  remainder  BH : 
in  the  same  manner  it  is  demonstrated,  that  BF  is  equal  to  CL : 
and  since  the  points  B,  D,  G,  F,  are  in  a  circle,  the  angle  DGF 
will  be  equal  to  the  exterior  and  opposite  angle  FBH,  (22.  3.)  ; 
■wherefore  their  halves  BGD,  HBK  will  be  equal  to  one  another: 
the  right  angled  triangles  BCiD,  HBK,  will  therefore  be  equi- 
angular, and  GD  will  be  to  BD,  as  BH  to  HK,  and  the  rectan- 


-488  PLANE  rRIGONOMETRY. 

gle  contained  by  GD,  HK  will  be  equal  to  the  rectangle  DBti 
or  BFC :  but  since  AH  is  to  HK,  as  AD  to  DG,  the  rectangle 
HAD  (22.  6.)  will  be  to  the  rectangle  contained  by  HK,  DG,  or 
the  rectangle  BFC,  (as  the  square  of  AD  is  to  the  square  of  DG, 
that  is)  as  the  square  of  the  radius  to  the  square  of  the  tangent 
of  the  angle  DAG,  that  is,  the  half  of  BAG :  but  HA  is  half 
the  perimeter  of  the  triangle  ABC,  and  AD  is  the  excess  of  the 
same  above  HD,  that  is,  above  the  base  BC  :  but  BV  or  CL  is 
the  excess  of  HA  or  AL  above  the  side  AC,  and  FC,  or  Hl3,  is 
the  excess  of  the  same  HA  above  the  side  AB ;  therefore  the 
;  rectangle  contained  by  half  the  perimeter,  and  the  excess  of  the 
same  above  the  base,  viz.  the  rectangle  HAD,  is  to  the  rectangle 
contained  by  the  straight  lines  by  which  the  half  of  the  peri- 
meter exceeds  the  other  two  sides,  that  is,  the  rectangle  BFC, 
as  the  square  of  the  radius  is  to  the  square  of  the  tangent  of  half 
the  angle  BAC  opposite  to  the  base.     Q.  E.  D. 


PROP.  VII.    FIG.  12, 15. 

IN  a  plane  triangle,  the  base  is  to  the  sum  of  the  sides, 
as  the  difference  of  the  sides  is  to  the  sum  or  difference 
of  tlie  segments  of  the  base  made  by  the  perpendicular 
upon  it  from  the  ^'ertex,  according  as  the  squai"C  of  the 
greater  side  is  greater  or  less  than  the  sum  of  the  squares 
of  the  lesser  side  and  the  base. 

Let  y\BC  be  a  plane  triangle ;  if  from  A  the  vertex  be  drawn 
a  straight  line  AD  perpendicular  upon  the  base  BC,  the  base  BC 
uill  be  to  the  sum  of  the  sides  BA,  AC,  as  the  difference  of  tire 
same  sides  is  to  the  sum  or  difference  of  the  segments  CD,  BD, 
according  as  the  square  of  AC  the  greater  side  is  greater  or 
less  than  the  sum  of  the  squares  of  the  lesser  side  AB,  and  the 
base  BC. 

About  A  as  a  centre,  -vvith  AC  the  greater  side  for  a  distance^ 
let  a  circle  be  described  meeting  AB  produced  in  E,  F,  and  CB 
in  G  :  it  is  manifest  that  FB  is  the  sum,  and  BE  the  difference 
of  the  sides;  and  since  AD  is  perpendicular  to  GC,  GD  CD 
will  be  equal  ;  consequently  GB  will  be  equal  to  the  sum  or  dif- 
ference of  the  segments  CD  BD,  according  as  the  perpendicular 
AD  meets  the  base,  or  the  base  produced  ;  that  is,  (by  conv.  12. 
and  13.  2.)  according  as  the  square  of  AC  is  greater  or  less  than 
the  sum  of  the  squares  of  AB,  BG  :  but  (by  35.  3.)  the  rectan-* 


PLANE  TRIGONOMETRY.  48g 

^le  CBG  is  equal  to  the  rectangle  EBF ;  that  is,  (16.  6.)  BC 
is  to  BF,  as  BE  is  to  BG  ;  that  is,  the  base  is  to  the  sum  of 
the  sides,  as  the  difference  of  the  sides  is  to  the  sum  or  differ- 
ence of  the  segments  of  the  base  made  by  the  perpendicular  from 
the  vertex,  according  as  the  square  of  the  greater  side  is  greater 
or  less  than  the  sum  of  the  squares  of  the  lesser  side  and  the 
base.     Q.  E.  D* 


tROP.  VIII.  PROB.    FIG.  14. 

^  The  sum  and  difference  of  two  magnitudes  being 
given,  to  find  them. 

Half  the  given  sum  added  to  half  the  given  difference,  Will  be 
the  greater,  and  half  the  difference  subtracted  from  half  the  sum^ 
will  be  the  less. 

For,  let  AB  be  the  given  sura,  AC  the  greater,  and  BC  the 
less.  Let  AD  be  half  the  given  snm  ;  and  to  AD,  DB,  which 
are  equal,  let  DC  be  added,  then  AC  will  be  equal  to  BD,  and 
DC  together ;  that  is,  to  BC,  and  twice  DC ;  consequently  twice 
DC  is  the  difference,  and  DC  half  that  difference  :  but  AC  the 
greater  is  equal  to  AD,  DC ;  that  is,  to  half  the  sum  added  to 
half  the  difference,  and  BC  the  less  is  equal  to  the  excess  of  BD, 
half  the  sum  above  DC  half  the  difference.     Q.  E.  F. 

SCHOLIUM. 

Of  the  six  parts  of  a  plane  triangle  (the  three  sides  and  three 
angles)  any  three  being  given^  to  find  the  other  three  is  the  bu- 
siness of  plane  trigonometry  ;  and  the  several  cases  of  that  pro- 
blem may  be  resolved  by  means  of  the  preceding  ^propositions, 
as  ill  the  two  following,  with  the  tables  annexed.  In  these,  the 
solution  is  expressed  by  a  fourth  proportional  to  three  given  lines  ; 
but  if  the  given  parts  be  expressed  by  numbers  from  trigonome- 
trical tables,  it  may  be  obtained  arithmetically  by  the  common 
Rule  of  Three. 


2<f^ote.  In  the  tables  the  foltowitig  abbreviations  are  used.  U  is  put  for  the 
Radius;  T  for  Tangent ;  and  S  for  Sine  Degrees,  minutes,  seconds,  kc.. 
are  written  in  this  manner;  30**  25'  13"  &c.  which  signifies  ;iO  degrees,  2* 
minutes,  13  secondr,,  &c-, 

3Q 


490 


PLANE  TRIGONOMETRY. 


SOLUTION    OF    THE    CASES  OF    RIGHT  ANGLED   TRIANGLES. 

GENERAL  PROPOSITION. 

IN  a  right  angled  triangle,  of  the  three  sides  and  three 
angles,  any  t'wo  being  given  besides  the  right  angle,  the 
other  three  may  be  found,  except  when  the  two  acute  an- 
gles are  i2:iven,  in  which  case  the  ratios  of  the  sides  ai'e 
only  given,  l)Cing  the  same  with  the  ratios  of  the  sines 
of  the  angles  opposite  to  them. 

It  is  manifest  from  47.  I.  that  of  the  two  sides  and  hypothe- 
ijusc  any  two  be  given,  the  third  may  also  be  found.  It  is  also 
manifest  from  32.  1.  that  if  one  of  the  acute  angles  of  a  right- 
angled  triangle  be  given,  the  other  is  also  given,  for  it  is  the 
complement  of  the  former  to  a  right  angle. 

If  two  angles  of  any  triangle  be  given,  the  third  is  also  given, 
being   the   supplement  of  the  two  given   angles   to  two    right 
angles. 
Fig.  15.        The  other  cases  may  be  resolved  by  help  of  the  preceding 
propositions,  as  in  the  following  table : 


S 
S  - 

V 

s 

\'' 

s 
s 

S  f 

s 

s 

S  / 
S  ' 

s 
s 
s 


SOUGHT. 


1  wo  sides,    Ai! 
AC. 


.li,  BC,  a  side  an( 
lu;  hypothci^ust . 


AB,  B,  a  sule  am, 
m  antrle. 


.vB  and  B,  a  siot 
:ind  an  ane;le. 


1  he  an- 
gles B,  C. 


i  he  au 
;les  B,  C. 


1  ne  other 
side  AC. 


AB  :  AC  :   :  K  :    1,  Jb.  ot 

which  C  is  the  complement. 


BC  :   BA  :   :  K  :   b,  C,  ot 
which  B  is  the  complement. 


K  :   T,  B  :   :  BA  :   AC 


The  hy-       S,  C  :   R  :   :  BA  :  BC. 
pothenust 
BC. 


BC  and  B,  tlie 
hypothenuseandan 
angle. 


1  he  side 
AC. 


K  :   b,  B  :   :  BC  :   CA. 


-^ 

*S 
S 

s 
s 
s 
s 
s 
s 
s 
s 
s 
s 

$ 

s 

\ 

s. 

\ 


These  five  cases  are  resolved  by  prop.  1. 


PLANE  TRIGONOMETRY. 


^9J 


SOLUTION  OF   THE    CASES    QT    OBLIQUE    ANGLED   TET ANGLES. 

GENERAL  PROPOSITION. 

IN  an  oblique  angled  triangle,  of  the  three  sides  and 
three  angles,  any  three  being  given,  tlie  other  three  ma\' 
be  found,  except  when  the  three  angles  are  given ;  in 
which  case  the  ratios  of  the  sides  are  onl}^  gi\en,  being 
the  same  with  the  ratios  of  tlie  sines  of  the  angles  oppo- 
site to  them. 


i 

s 
s 


,*,  ii,  a/iti  tiicrc- 
forc  C,  and  the  sidt 


>o, 


AC. 


Aii,  AC,  unci  ii 
wo  sides  and  anar.- 
^le  opposite  to  one 
if  them. 


ihe  un- 
rles  A  anci 


Ali,  AC,  ak.d  A, 
two  sides  and  tht 
included  angle. 


'ilic  all- 
eles B  and 
C. 


*S 

Bc,  J; 


S,  C  ;   8,  A  :  :  AB 
and  also,  S,  C  :  S,  B,  :  :  AB  S 

AC.     (2.) \ 

AC  :   Ab  :   :  b,  B  :  b,  C  ^J 

(2.)     This   case  admits  of  S 

Lwo  solutions  ;  tor  C  may  be  Ij 

reuter  or  icss  than  a  quad-  S 

int.  (Cor.  to  def.  4.)  ^^ 


-vB  +  AC  :  aB— AC 
C-t-B    :    T,    C— B    : 


Fig.  16, 
17. 


(3.):j 

2  2  <) 

the  sum  and  difference  of  ^ 
the  angles  C,  B  being  given,  S 
each  of  them  is  given.  (7.)  «, 
Othervjine.   Vig.  18.  S 

BA  :  AC  :  :  R  :  T,  ABC,  $ 
and  also  R  :  T,  ABC— 45"  S 
!  T,  B+C  ;  r.  B— C  :  (4,)  v 

2  Z  ^ 

therefore  B  and  C  are  given 
ds  before.     (7.) 


^- 


/-/"  r^  f  /^j~y'^y. 


\ 

s 

s 
s 

s 


PLANE  TRIGONOMETRY. 


S 

s 

:: 

s 

s 
s 

S 


GIVEN. 


SOUGHT. 


s 


th 


AB,  BC,  CA  the 

ree  sides. 


A,  B,  C. 

the    three 
ingles. 


-2  ACx^b  :  ACrH-vB.y — 
vBy:  :R:CoS,C.  If  AB? 
+  CB5r  be  greater  than  ABq. 

ig.  16. 
2  ACxCB  :  ABy— ACy 
— CBy  :  :  R  :  Co  S,  C  If 
ABy  be  greater  than  ACq-\- 
Ciiq.  Fig.  17.  (4.) 
Otherwise, 

Let  AB+BC+AC=2P. 


!'Xl    —  /^B  :  P  —  AC    X 


P— liC  :  :  Rg  :  Ty,  -|  C,  and 
iience  C  is  known.  (5.) 
Otherivise. 

Let  AD  be  perpendicular 
io  BC.  )..  If  AB^  be  less 
than  ACy.+CBy.  Fig.  16. 
BC  :  BA+AC  :  :  BA— AC 
:  BD^DC,  and  BC  the  sum 
of  BD,  DC  is  given  ;  there- 
fore each  of  them  is  given. 
•(7.) 

2.  If  ABy  be  greater  than 
VCy+CBfy.  Fig.  17.  BC  : 
BA  +  AC  :  :  BA— AC  :  BD 
+DC;andBCthedifference 
of  BD,  DC  is  given,  there- 
fore each  of  them  is  given. 
(7.) 

And  CA  :  CD  :  :  R  :  Co 
S,  C.  (1.)  and  C  being 
found,  A  and  B  are  found 
Dv  case  2.  or  3. 


S 

^ 

s 
s 
s 
s 
s 
s 
s 

I 

s. 
s 

^ 

s 
s 
s. 
s 

s 

s 

s 
<» 

s 

s 

s 
s 
s 
s 
s 
s 
s 
s 
s 
s 
s 
s 
s 


SPHERICAL  TRIGONOMETRY, 


DEFINITIONS. 


I. 


X  HE  pole  of  a  circle  of  the  sphere  is  a  point  in  the  superficies 
of  the  sphere,  from  which  all  straight  lines  drawn  to  the  cir- 
cumference of  the  circle  are  equal. 


ir. 

A  great  circle  of  the  sphere  is  any  whose  plane  passes  through 
the  centre  of  the  sphere,  and  whose  centre  therefore  is  the 
same  with  that  of  the  sphere. 

lU. 

A  spherical  triangle  is  a  figure  upon  the  superficies  of  a  sphere 
comprehended  by  three  arches  of  three  great  circles,  each  of 
which  is  less  than  a  semicircle. 

IV. 

A  spherical  angle  is  that  which  on  the  superficies  of  a  sphere  is 
contained  by  two  arches  of  great  circles,  and  is  the  same  with 
the  inclination  of  the  planes  of  these  great  circles. 

PROP.  I. 

GREAT  circles  bisect  one  another. 

As  they  have  a  common  centre  their  common  section  will  be 
A  diameter  of  each  which  will  bisect  them. 

PROP.  11.   FIG.  I. 

THE  arch  of  a  great  circle  bet\^ixt  the  pole  and  the 
circumference  of  another  is  a  quadrant. 

Let  ABC  be  a  great  circle,  and  D  its  pole:  if  a  great  circle 
DC  pass  through  D,  and  meet  ABC  in  C,  the  arch  DC  will  be 
a  quadrant. 

Let  the  great  circle  CD  meet  ABC  again  in  A,  and  let  AC  be 
the  coroaion  section  of  the  great  circles,  which  will  pass  through 


49'4.  SPHERICAL  TRIGONOMETRY. 

E  the  cQhtre  of  the  sphere:  join  DE,  DA,  DC  :  by  def.  1.  DA, 
DC  are  equal,  and  AE,  EC  are  also  equal,  and  DE  is  common ; 
therefore  (8.  1.)  the  angles  DEA,  DEC  are  equal ;  wherefore  the 
arches  DA,  DC  are  equal,  and  consequently  each  of  them  is  ^ 
quadrant.     Q.  E.  D. 


PROP.  III.    FIG.  2. 

IF  a  sfreat  circle  be  described  meetinf^r  two  ^reat  cir- 
cics  AB,  AC  passing  through  its  pole  A  in  B,  C,  the 
angle  at  the  centre  of  the  sphere  upon  the  circumierence 
BC,  is  the  same  with  the  sphericd  angle  BAC,  and  the 
arch  BC  is  called  the  measure  of  the  spherical  ansrle 
BAC. 


Let  the  planes  of  the  great  circles  AB,  AC  intersect  one  an- 
other in  the  straight  line  AD  passing  through  D  their  common 
centre ;  join  DB,  DC. 

Since  A  is  the  pole  of  BC,  AB,  AC  will  be  quadrants,  and  the 
angles  ADB,  ADC  right  angles  ;  therefore  (6.  def.  1 1.)  the  ai)gle 
CBD  is  the  inclination  of  the  planes  of  the  circles  AB,  AC;  that 
is,  (def.  4.)  the  spherical  angle  BAC.     Q.  E.  D. 

CoR.  If  through  the  point  A,  two  quadrants  AB,  AC  be  drawn, 
the  point  A  will  be  the  pole  of  the  great  circle  BC,  passing  through 
their  extremities  B,  C. 

Join  AC,  and  draw  AE  a  straight  line  to  any  other  point  E  in 
BC  ;  join  DE  :  since  AC,  AB  are  quadrants,  the  angles  ADB, 
ADC  arc  right  angles,  and  AD  will  be  perpendicular  to  the  plane 
of  BC  :  therefore  the  angle  ADE  is  a  right  angle,  and  AD,  DC 
are  equal  to  AD,  DE,  each  to  each  ;  therefore  AE,  AC  are  equal, 
and  A  is  the  pole  of  BC,  by  def.  1.     Q.  E.  D. 


PROP.  IV.     FIG.  3. 

IN  isosceles  spherical  triangles,  the  angles  at  the  base 

are  equal. 

Let  ABC  be  an  isosceles  triangle,  and  AC,  CB  the  equal  sides  ; 
the  angles  BAC,  ABC,  at  the  bai;e  AB,  are  equal. 

Let  D  be  the  centre  of  the  sphere,  and  join  DA,  DB, 
DC;  in  DA  Vake  any  point- E,  from  which  draw,  in  the  plane 
\DC3  the  strujght  line  EF  at  right  angles  to  ED  needing  CD 


SPHERICAL  TRIGONOMETRY*  -fRS 

ill  F,  and  draw,  in  the  plane  ADB,  EG  at  right  angles  to  the 
same  ED;  therefore  the  rectilineal  angle  FEG  is  (6.  clef.  11.) 
the  inclination  of  the  planes  ADC,  ADB,  and  therefore  is  the 
same  with  the  spherical  angle  BAC  ;  from  F  draw  FH  perpendicu- 
lar to  DB.  and  from  H  draw,  in  the  plane  ADB,  the  straight  tine 
HG  at  right  angles  to  HD  meeting  EG  in  G,  and  join  GF.  Be- 
cause DE  is  at  right  angles  to  EF  and  EG,  it  is  perpendicular  to 
the  plane  FEG,  (4.  1 1.)  and  therefore  the  plane  FEG,  is  perpen- 
dicular to  the  plane  ADB,  in  which  DE  is  :  (18.  1 1.)  in  the  same 
manner  the  plane  FHG  is  perpendicular  to  the  plane  ADB;  and 
therefore  GF  the  common  section  of  the  planes  FEG,  FHG  is 
perpendicular  to  the  plane  ADB  ;  (19.  11.)  and  because  the  an- 
gle FHG  is  the  inclination  of  the  planes  BDC,  BDA,  it  is  the 
same  with  the  spherical  angle  ABC;  and  the  sides  AC,CBofthe 
spherical  triangle  being  equal, the  angles  EDF,  HDF,  which  stand 
upon  them  at  the  centre  of  the  sphere,  are  equal ;  and  in  the 
triangles  EDF,  HDFj  the  side  DF  is  common,  and  the  angles 
DEF,  DHF  are  right  angles;  therefore  EF,  FH  are  equal;  and 
in  the  triangles  FEG,  FHG  the  side  GF  is  common,  and  the 
sides  EG,  GH  will  be  equal  by  the  47.  Land  therefore  the  angle 
FEG  is  equal  to  FHG  ;  (8.  1.)  that  is,  the  spherical  angle  BAC 
is  equal  to  the  spherical  angle  ABC. 


PROP.  V.    FIG.  3. 

IF,  in  a  spherical  triangle  ABC,  two  of  the  angles 
BAC,  ABC  be  equal,  the  sides  BC,  AC  opposite  to 
them,  are  equal. 

Read  the  constructloii  and  demonstration  of  the  preceding- 
proposition,  unto  the  words,  "  and  the  sides  AC,  CB,"  &cc.  and 
the  rest  of  the  demonstration  will  be  as  follows,  viz. 

And  the  spherical  angles  BAC,  ABC,  being  equal,  the  recti- 
lineal angles  FEG,  FHG,  which  are  the  same  "with  them,  are 
equal ;  and  in  the  triangles  FGE,  FGHthe  angles  at  G  are  right; 
angles,  and  the  side  FG  opposite  to  two  of  the  equal  angles  is 
coti.mon  ;  therefore  (26.  1.)  EF  is  equal  to  FH  ;  and  in  the  right 
angled  triangles  DEF,  DHF  the  side  DF  is  common  ;  wherelore 
(47.  1.)  ED  is  equal  to  DH,  and  the  angles  EDF,  HDF,  are 
therefore  equal.  (4.  1.)  and  consequently  the  sides  AC,  RCof  tht; 
spherical  triangle  are  equal. 


49«  SPHERICAL  TRIGONOMETRY. 


PROP.  VI.    FIG.  4. 

ANY  two  sides  of  a  spherical  triangle  are  greater  than 
the  tliird. 

Let  ABC  be  a  spherical  triangle,  any  two  sides  AB,  BC  wili 
be  greater  than  the  other  side  AC. 

Let  D  be  the  centre  of  the  sphere ;  join  DA,  DB,  DC. 

The  solid  angle  at  D,  is  contained  by  three  plane  angles,  ADB, 
ADC,  BDC  ;  and  by  20.  11.  any  two  of  them  ADB,  BDC  are 
greater  than  the  third  ADC  ;  that  is,  any  two  sides  AB,  BC  of 
the  spherical  triangle  ABC,  are  greater  than  the  third  AC. 


PROP.  VII.     FIG.  4. 

THE  three  sides  of  a  spherical  triangle  are  less  than  a 
circle. 

Let  ABC  be  a  spherical  triangle  as  before,  the  three  sides  A&j 
BC,  AC  are  less  than  a  circle. 

Let  D  be  the  centre  of  the  sphere :  the  solid  angle  at  D  is 
contained  by  three  plane  angles  BDA,  BDC,  ADC,  which  to- 
gether are  less  than  four  right  angles,  (^l.  11.)  therefore  the 
sides  AB,  BC,  AC  together,  will  be  less  than  four  quadrants;  that 
rs,  less  than  a  circle. 


PROP.  VIIL    FIG.  5. 

IN  a  spherical  triangle  the  greater  angle  is  opposite  to 
the  greater  side;  and  conversely. 

Let  ABC  be  a  spherical  triangle,  the  greater  angle  A  is  op- 
posed to  the  greater  side  BC. 

Let  the  angle  Bx\D  be  made  equal  to  the  angle  B,  and  then 
BD,  DA  will  be  equal,  (5.  of  this)  and  therefore  AD,  DC  are 
equal  to  BC  ;  but  AD,  DC  are  greater  than  AC,  (6.  of  this), 
therefore  BC  is  greater  than  AC,  that  is,  the  greater  angle  A  is 
opposite  to  the  greater  side  BC.  The  converse  is  demonstrated 
as  prop.  19.  I.  El.    Q.  li.D. 


SPHERICAL  TRIGONOMETRY.  497 


PROP.  IX.     FIG.  6. 

IN  any  spherical  triangle  ABC,  if  the  sum  of  the  sides 
AB,  BC,  be  greater,  equal,  or  less  than  a  semicircle, 
the  internal  angle  at  the  base  AC  will  be  greater,  equal, 
or  less  than  the  external  and  opposite  BCD  ;  and  there- 
fore the  sum  of  the  angles  A  and  ACB  will  be  greater, 
equal,  or  less  than  two  right  angles. 

Let  AC,  AB  produced  meet  in  D. 

1.  If  AB,  BC  be  equal  to  a  semicircle,  that  is,  to  AD,  JBC, 
BD  will  be  equal,  that  is,  (4.  of  this)  the  angle  D,  or  the  angle 
A  will  be  equal  to  the  angle  BCD. 

2.  If  AB,  BC  together  be  greater  than  a  semicircle,  that  is, 
greater  than  ABD,  BC  will  be  greater  than  BD  ;  and  therefore 
(8.  of  this)  the  angle  D,  that  is,  the  angle  A,  is  greater  than  the 
angle  BCD. 

3.  In  the  same  manner  it  is  shown,  that  if  AB,  BC  together 
be  less  than  a  semicircle,  the  angle  A  is  less  than  the  angle  BCD. 
And  since  the  angles  BCD,  BCA  are  equal  to  two  right  angles, 
if  the  angle  A  be  greater  than  BCD,  A  and  ACB  together  will 
be  greater  than  two  right  angles.  If  A  be  equal  to  BCD,  A  and 
ACB  together  will  be  equal  to  two  right  angles ;  and  if  A  be 
less  than  BCD,  A  and  ACB  will  be  less  than  two  right  angles. 
Q.  E.  D. 


PROP.  X.  FIG.  r. 

•  IF  the  angular  points  A,  B,  C  of  the  spherical  trian- 
gle ABC  be  the  poles  of  three  great  circles,  these  great 
circles  by  their  intersections  will  form  another  triangle 
FDE,  which  is  called  supplemental  to  the  former ;  that 
is,  the  sides  FD,  DE,  EF  are  the  supplements  of  the 
measures  of  the  opposite  angles  C,  B,  A,  of  the  trian- 
gle ABC,  and  the  measures  of  the  angles  F,  D,  E  of  the 
triangle  FDE,  will  be  the  supplements  of  the  sides  AC. 
BC,  BA,  in  the  triangle  ABC. 

3  R 


SPHERICAL  TRIGONOMETRY. 

Let  AB  produced  meet  DE,  EF  in  G,  M,  and  AC  meet  FD, 
FE  in  K,  L,  and  BC  meet  FD,  DE  in  N,  H. 

Since  A  is  the  pole  of  FE,  and  the  circle  AC  passes  through 
A,  EF  will  pass  through  the  pole  of  AC,  (13.  Ifi.  1.  rh.)  and 
since  AC  passes  through  C,  the  pole  of  FD,  FD  will  pass  through 
the  pole  of  AC ;  therefore  the  pole  of  AC  is  in  the  point  F,  in 
which  the  arches  DF,  EF  intersect  each  other.  In  the  same 
manner  D  is  the  pole  of  BC,  and  E  the  pole  of  AB. 

And  since  F,  E  are  the  poles  of  AL,  AM,  FL  and  EM  are 
quadrants,  and  FL,  EM  together,  that  is,  FE  and  ML  together, 
:ire  equal  to  a  semicircle.  But  since  A  is  the  pole  of  ML,  ML 
is  the  measure  of  the  angle  BAC,  consequently  FE  is  the  sup- 
plement of  the  measure  of  the  angle  BAC.  In  the  same  man- 
ner, ED,  DF  are  the  supplements  of  the  measures  of  the  angles 
ABC,  BCA. 

Since  likewise  CN,  BH  are  quadrants,  CN,  BH  together,  that 
is,  NH,  BC  together  are  equal  to  a  semicircle;  and  since  D  is 
the  pole  of  NM,  NH  is  the  measure  of  the  angle  FDh,  there- 
fore the  measure  of  the  angle  FDE  is  the  supplement  of  the  side 
BC.  In  the  same  manner,  it -is  shown  that  the  measures  of  the 
angles  DEF,  EFD  are  the  supplements  of  the  sides  AB,  AC,  ii> 
the  triangle  ABC.     Q.  E.  D. 


PROP.  XL    FIG.  r. 

THE  three  angles  of  a  spherical  triangle  are  greater 
than  two  right  angles,  and  less  than  six  right  angles. 

The  measures  of  the  angles  A,  B,  C,  in  the  triangle  ABC, 
together  with  the  three  sides  of  the  supplemental  triangle  DEF, 
are  (10.  of  this)  equal  to  three  semicuiles;  but  the  three  sides 
•of  the  triangle  FDE,  are  (7.  of  this)  less  than  two  semicircles; 
therefore  the  measures  of  the  angles  A,  B,  C  are  greater  than 
a  semicircle ;  and  hence  the  angles  A,  B,  C  are  greater  than  two 
right  angles. 

All  the  external  and  internal  angles  of  any  triangle  are  equal 
to  six  right  angles ;  therefore  all  the  internal  angles  are  less  than 
six  right  angles. 


PROP.  XIL    FIG.  8. 

IF  from  any  point  C,  which  is  not  the  pole  of  the  great 
cbxie  AjjD,  there  be  drawn  arches  of  great  circles  CA, 


SPHERICAL  TRIGONOMETRY.  499 

CD,  CE,  CF,  &c.  the  greatest  of  these  is  CA,  which 
passes  througli  H  the  pole  o'l  ABD,  and  CB  the  remain- 
der of  ACB  is  the  least,  and  of  any  others  CD,  CE,  CF, 
&c.  CD,  which  is  nearer  to  CA,  is  greater  than  CE, 
which  is  more  remote. 

Let  the  common  section  of  the  planes  of  the  great  circles  ACB, 
ADB  be  AB:  and  from  C,  draw  CG  perpendicular  to  AB,  which 
will  also  be  perpendicular  to  the  plane  ADB  (4.dcf.  11.);  join  GD, 

GE,  GF,  CD,  CE,  CF,  CA,  CB. 

Of  all  the  straight  lines  (Iruwn  from  G  to  the  circumference 
ADU,  GA  is  the  greatest,  and  GB  the  least  (7.  3.);  and  GD 
vil/u  h  is  tuartr  lo  GA  is  greater  than  GE,  which  is  more  re- 
mole.  Tiu-  triangles  CGA,  CGD  are  right  angled  at  G,  and  they 
have  the  comniou  siue  CG  ;  therefore  the  squares  of  CG,  GA 
togithi.-r,  tiiat  is.  the  square  of  CA,  is  greater  than  the  squares 
of  t'G.  GD  together  that  is,  the  square  of  CD  ;  and  CA  is  great- 
er tiidii  CU,  and  thevi-fore  the  arch  CA  is  greater  than  CD.  In 
tlie  same  manner,  since  GD  is  greater  than  GE,  and  GE  than 

GF,  ix-c.  it  is  shown  tiiat  CD  is  grealer  than  CE,  and  CE  than 
CF,  8cc.  and  coiiseqv.enlly,  the  arch  CD  greater  than  the  arch 

CE,  and  the  arch  CE  greater  than  the  arch  CF,  &c.  And  since 
GA  is  the  greatest,  and  GB  the  least  of  all  the  straight  lines 
drawn  from  G  to  the  circumference  ADB,  it  is  manifest  that  CA 
is  the  greatest,  and  Cii  the  least  of  all  the  straight  lines  drawn 
fro  ■.  I.'  to  the  circumference  ;  and  therefore  the  arch  CA  is  the 
gretttesi,  and  CB  tiie  least  of  all  the  circles  drawn  through  C, 
meeting  ADB.     Q.  E.  D. 


PROP.  Xin.     FIG.  9. 

IN  a  right  angled  spherical  triangle  the  sides  arc  of  tli^ 
same  affection  v*  ith  the  opposite  angles ;  that  is,  if  the 
sides  be  greater  or  less  than  quadrants,  the  opposite  an- 
gles w^iil  be  greater  or  less  than  right  angles. 

Let  ABC  be  a  spherical  triangle  right  angled  at  A,  any  side 
AB,  will  be  of  the  same  affection  with  the  opposite  angle  ACB. 

Case  1.  Let  AB  be  less  than  a  quadrant,  kl  AE  be  a  quadrant, 
and  let  EC  be  a  great  circle  passing  through  E,  C.  Since  A  is 
a  right  angle,  and  AE  a  quadrant,  li  is  the  pole  of  the  great  cir-. 


5Q®  SPHERICAL  TRIGONOMETRY. 

cle  AC,  and  ECA  a  right  angle :  but  EC  A  is  greater  than  BCAj 
therefore  BCA  is  less  than  a  right  angle.  Q.  E.  D. 
•fig.  10.  Case  3'  Let  AB  be  greater  than  a  quadrant,  make  AE  a  quad- 
rant, and  let  a  great  circle  pass  through  C,  E,  ECA  is  a  right 
angle  as  before,  and  BCA  is  greater  than  ECA,  that  is,  greater 
than  a  right  angle.     Q.  E.  D. 

PROP.  XIV. 

IF  the  two  sides  of  a  right  angled  spherical  triangle  be 
of  the  same  affection,  the  hypothenuse  will  be  less  than 
a  quadrant ;  and  if  they  be  of  different  affection,  the  hy- 
pothenuse will  be  greater  than  a  quadrant. 

Let  ABC  be  a  right  angled  spherical  triangle,  if  the  two  sides 

AB,  AC  be  of  the  same  or  of  different  affection,  the  hypothe- 
nuse BC  will  be  less  or  greater  than  a  quadrant. 

Fig.  9.  Case  1 .  Let  AB,  AC  be  each  less  than  a  quadrant.  Let  AE, 
AG  be  quadrants  ;  G  will  be  the  pole  of  AB,  and  E  the  pole  of 

AC,  and  EC  a  quadrant;  but,  by  prop.  12.  CE  is  greiter  than 
CB,  since  CB  is  farther  off  from  CGD  than  CE.  In  the  same 
manner,  it  is  shown  that  CB,  in  the  triangle  CBD,  where  the 
two  sides  CD,  BD  are  each  greater  than  a  quadrant,  is  less  than 
CE,  that  is,  less  than  a  (quadrant.     Q  E.  D. 

Fig.  10,  Case  2.  Let  AC  be  less,  and  AB  greater  than  a  quadrant;  then 
the  hypothenuse  BC  will  be  greater  than  a  quadrant ;  for  let  AE 
be  a  quadrant,  then  E  is  the  pole  of  AC,  and  EC  will  be  a 
quadrant.  But  CB  is  greater  than  CE  by  prop.  12.  since  AC 
passes  through  the  pole  of  ABD.     Q.  E.  D. 


PROP.  XV. 

IF  the  hypothenuse  of  a  right  angled  triangle  be  great- 
er or  less  than  a  quadrant,  the  sides  will  be  of  different 
or  the  same  affection. 

This  is  the  converse  of  the  preceding,  and  demonstrated  in  the 
same  manner. 


SPHERICAL  TRIGONOMETRY.  504 


PROP.  XVI. 

In^  any  spherical  triangle  ABC,  if  the  perpendicu- 
lar AD  from  A  upon  the  base  BC,  fall  within  the  tri- 
angle, ,  the  angles  B  and  C  at  the  base  will  be  of  the 
same  affection ;  and  if  the  perpendiculai'  fall  without 
the  triangle,  the  angles  B  and  C  will  be  of  diil'erent 
affection. 

1.  Let  AD  fall  within   the  triangle;    then  (13.  of  this)  since  Fig.  11. 
ADB,  ADC  are  right  angled   spherical   triangles,  the  angles  B, 

C  must  each  be  of  the  same  aflection  as  AD. 

2.  Let  AD   fall  without   the  triangle,  then  (13.    of  this)  the  Fig.  12. 
angle    B  is  of  the  same  affection    as  AD  ;  and  by  the  same  the 
angle  ACD  is  of  the  same  affection  as  AD  ;  therefore  the  angle 
ACB  and  AD  are  of   different  affection,  and  the  angles  B  and 
ACB  of  different  affection. 

CoR.  Hence  if  the  angles  B  and  C  be  of  the  same  affection, 
the  perpendicular  will  ftill  within  the  base  ;  for,  if  it  did  not 
(16.  of  this),  B  and  C  would  be  of  different  affection.  And  if 
the  angles  B  and  C  be  of  opposite  affection,  the  perpendicular 
■will  fall  without  the  triangle;  for,  if  it  did  not  (16.  of  this), 
the  angles  B  and  C  would  be  of  the  same  affection,  contrary  to 
the  supposition. 


PROP.  XVIL  FIG.  13. 

IN  right  angled  spherical  triangles,  the  sine  of  ei- 
ther of  the  sides  about  the  right  angle,  is  to  the  radius 
of  the  sphere,  as  the  tangent  of  the  remaining  side 
is  to  the  tangent  of  the  angle   opposite  to  that  side. 

Let  ABC  be  a  triangle,  having  the  right  angle  at  A  ;  and 
let  AB  be  either  of  the  sides,  the  sine  of  the  side  AB  will  be 
to  the  radius,  as  the  tangent  of  the  other  side  AC  to  the  tan- 
gent of  the  angle  ABC,  opposite  to  AC  Let  D  be  the  cen- 
tre of  the  sphere  ;  join  AD,  BD,  CD,  and  let  AE  be  drawn 
perpendicular  to  BD,  which  therefore  will  be  the  sine  of  the 
arch  AB,  and  from  the  point  E,  let  there  be  drawn  in  the 
plane  BDC  the  straight  line  EF  at  right  angles  to  BD,  meeting 
DC  in  E,  and  let  AF  be  joined.  Since  therefore  the  straiglit 
line  BE  is  at    ri^ht  angles  to  both   EA  and  EF,  it  will  also  be 


|02  SPHERICAL  TRIGONOMETRY. 

at  right  angles  to  the  plane  AEF  (4,  11.)  wherefore  the  plane 
ABD,  which  passes  through  DE,  is  perpendicular  to  the  plane 
AEF  (18.  11.),  and  the  plane  AEF  pt  rpendicular  to  x\BD  :  the 
plane  ACD  or  AFD  is  also  perpendicular  to  the  same  ABD: 
therefore  the  common  section,  viz.  the  straight  line  AF,  is  at 
right  angles  to  the  plane  ABD  (19.  II.):  and  FAE,  FAD  are 
right  angles  (3.  def.  11.);  therefore  AF  is  the  tangent  ol  the 
arch  AC;  and  in  the  rectilineal  triangle  AEF,  having  aright' 
angle  at  A,  AE  will  be  to  the  radius  as  AF  to  the  tangent  of  the 
angle  AEF  (1.  PI.  Tr.) ;  but  AE  is  the  sine  of  the  arch  AB, 
and  AF  the  tangent  of  the  arch  AC,  and  the  angle  AEF  is  the 
inclination  of  the  planes  CBD,  ABD  (6.  def.  11.),  or  the  sphe- 
rical angle  ABC  :  therefore  the  sine  of  the  arch  AB  is  to  the 
radius  as  the  tangen*,  of  the  arch  AC,  to  the  tangent  of  the 
opposite  angle  ABC. 

CoR.  1.  If  therefore  of  the  two  sides,  and  an  angle  oppo- 
site to  one  of  them,  any  two  be  given,  the  third  will  also  be 
given. 

CoR.  2.  And  since  by  this  proposition  the  sine  of  the  side 
AB  is  to  the  radius,  as  the  tangent  of  the  other  side  AC  to  the 
tangent  of  the  angle  ABC  opposite  to  that  side ;  and  as  the 
radius  is  to  the  co-tangent  of  the  angle  ABC,  so  is  the  tangent 
of  the  same  angle  ABC  to  the  radius  (Cor.  2.  def.  PI.  Tr.), 
by  equality,  the  sine  of  the  side  AB  is  to  the  co-tangent  of  the 
angle  ABC  adjacent  to  it,  as  the  tangent  of  the  other  side  AC 
to  the  radius. 


PROP.  XVIII.    FIG.  13. 

IN  right  angled  spherical  triangles  the  sine  of  the 
hypothenusc  is  to  the  radius,  as  the  sine  of  either 
side  is  to  the  sine  of  the  angle  opposite   to  that  side. 

Let  the  triangle  ABC  be  right  angled  at  A,  and  let  AC  be 
either  of  the  sides;  the  sine  of  the  hypnthenuse  BC  will  be  to 
the  radius  as  the  sine  of  the  arch  AC  is  to  the  sine  of  the  angle 
ABC. 

.  Let  D  be  the  centre  of  the  sphere,  and  let  CG  I)e  drawn  per- 
pendicular to  DB,  which  will  therefore  be  the  sine  of  the  hy- 
pothenusc BC  ;  aad  from  the  point  G  let  there  be  drawn  in  the 
plane  ABD  the  straight  line  GM  perpendicular  to  DB,  and  let 
CH  be  joined  :  CH  will  be  at  right  angles  to  the  plane  ABD,  as 
was  shown  in  the  preceding  proposition  of  the  straight  line 
FA:  wherefore  CHD,  CHG  are  right  angles,  and  CH  is  the 
sine  of  the  arch  AC  ;  and  in  the  triangle  CHG,  having  the  right 


SPHERICAL  TRIGONOMETRY.  50! 

angle  CHG,  CG  is  to  the  radius  as  CH  to  the  sine  of  the  angle 
CGH  (1  PI  Tr.):  but  since  CG,  HG  are  at  right  angles  to 
DGB,  wliich  is  the  common  section  of  the  planes  CBD,  ABD, 
the  ani>!e  CGH  will  be  equal  to  the  inclination  of  these  planes 
(6.  clef.  11.);  that  is,  to  the  spherical  angle  ABC.  The  sine, 
therefore,  of  the  hypothenuse  CB  is  to  the  radius  as  the  sine  of 
the  side  AC  is  to  the  sine  of  the  opposite  angle  ABC.  Q.  E.  D. 
Cor.  Of  theSf  three,  viz.  the  hypothenuse,  a  side,  and  the  an- 
gle opposite  to  that  side,  any  two  being  given,  the  third  is  also 
given  by  prop.  2. 


PROP.  XIX.    FIG.  14. 

IN  right  angled  spherical  triangles,  the  co-sine  of  the 
hypothenuse  is  to  the  radius  as  the  co-tangent  of  either  of 
the  angles  is  to  the  tangent  of  the  remaining  angle. 

Let  ABC  be  a  spherical  triangle,  having  a  right  angle  at  A, 
the  co-sine  of  the  hypothenuse  BC  will  be  to  the  radius  as  the 
co-tangent  of  the  angle  ABC  to  the  tangent  of  the  angle   ACB. 

Describe  the  circle  DE,  of  which  B  is  the  pole,  and  let  it  meet 
AC  in  F,  and  the  circle  BC  in  E  ;  and  since  the  circle  BD  pas- 
ses through  the  pole  B  of  the  circle  DF,  DF  will  also  pass  through 
the  pole  of  BO  (13.  18.  1.  Theod.  Sph.).  And  since  AC  is  per- 
pendicular to  BD,  AC  will  also  pass  through  the  pole  of  BD  ; 
wherefore  the  pole  of  the  circle  BD  will  be  found  in  the  point 
where  the  circles  AC,  DE  meet,  that  is,  in  the  point  F  :  the 
arches  FA,  FD  are  therefore  quadrants,  and  likewise  the  arches 
BD,  BE:  in  the  triangle  CEF,  rigiu  angled  at  the  point  E,  CE 
is  the  csmplement  of  the  hypothenuse  BC  of  the  tri;ingle  ABC, 
EF  is  the  complement  of  the  arch  ED,  which  is  the  mtasure  of 
the  angle  ABC,  and  FC  the  hypothenuse  of  tl^.e  triangle  CEF,  is 
the  complement  of  AC,  and  the  arch  AD,  vvhicii  is  the  nuusure 
of  the  anglt  CFE,  is  the  complement  of  AB. 

But  (17.  of  this)  in  the  triangle  CEF,  the  sine  of  the  side  CE 
is  to  the  radius,  as  the  tangent  of  the  other  side  i--  to  the  tan^'ent 
of  the  angle  ECF  opposite  to  it,  that  is,  in  the  triangle  ABC, 
the  co-sine  of  the  h>  pothenuse  BC  is  to  the  radius,  ai  the  co- 
tangent of  the  angle  ABC  is  to  the  tangent  of  t';e  angle  ACB. 
Q.  E.  D. 

CoR.  1.  Of  these  three,  viz.  the  hypothenuse  ai.  i  the  two  au- 
?jles,  any  two  being  given,  the  third  will  also  be  given. 


^04  SPHERICAL  TRIGONOMETRY. 

Cor.  2.  And  since  by  this  proposition  the  co-sine  of  the  hy- 
pothenuse  BC  is  to  the  radius  as  the  co-tangent  of  the  angle  ABC 
to  the  tangent  of  the  angle  ACB.  But  as  the  radius  is  to  the 
co-tangent  of  the  angle  ACB,  so  is  the  tangent  of  the  same  to 
the  radius  (Cor.  2.  def.  PI.  Tr.) ;  and,  ex  xquo,  the  co-sine  of 
i  the  hypothenuse  BC  is  to  the  co-tangent  of  the  angle  ACB,  as 

the  co-tangent  of  the  angle  ABC  to  the  radius. 


PROP.  XX.     FIG.  14. 

IN  right  angled  spherical  triangles,  the  co-sine  of  an 
angle  is  to  the  radius,  as  the  tangent  of  the  side  adjacent 
to  that  angle  is  to  the  tangent  of  the  hypothenuse. 

The  same  construction  remaining;  in  the  triangle  CEF  (17. 
of  this),  the  sine  of  the  side  EF  is  to  the  radius,  as  the  tangent 
of  the  other  side  CE  is  to  the  tangent  of  the  angle  CFE  opposite 
to  it  ;  that  is,  in  the  triangle  ABC,  the  co-sine  of  the  angle  ABC 
is  to  the  radius  as  (the  co-tangent  of  the  hypothenuse  BC  to  the 
co-tangent  of  the  side  AB,  adjacent  to  ABC,  or  as)  the  tangent 
of  the  side  AB  to  the  tangent  of  the  hypothenuse,  since  the  tan- 
gents of  two  arches  are  reciprocally  pi-oportional  to  their  co-tan- 
gents.    (Cor.  1.  def.  PI.  Tr.) 

CoR.  And  since  by  this  proposition  the  co-sine  of  the  angle 
ABC  is  to  the  radius,  as  the  tangent  of  the  side  AB  is  to  the  tan- 
gent of  the  hypothenuse  BC ;  and  as  the  radius  is  to  the  co-tan- 
gent of  BC,  so  is  the  tangent  of  BC  to  the  radius  ;  by  equality, 
the  co-sine  of  the  angle  ABC  will  be  to  the  co-tangent  of  the  hy- 
pothenuse BC,  as  the  tangent  of  the  side  AB,  adjacent  to  the  an^ 
gle  ABC  to  the  radius. 


PROP.  XXI.     FIG.  14. 

IN  right  angled  spherical  triangles,  the  co- sine  of  either 
of  the  sides  is  to  the  radius,  as  the  co-sine  of  the  hypo- 
thenuse is  to  the  co-sine  of  the  other  side. 

The  same  construction  remaining ;  in  the  triangle  CEF,  the 
sine  of  the  hypothenuse  CF  is  to  the  radius,  as  the  sine  of  the 
side  CE  to  the  sine  of  the  opposite  angle  CFE  (18.  of  this)  ; 
that  is,  in  the  triangle  ABC  the  co-sine  of  the  side  CA  is  to  the 


SPHERICAL  TRIGONOMETRY.      .  505 

radius  as  the  co-sine  of  the  hypothenuse  BC  to  the  co-sine  of  the 
other  side  BA.     Q.  E.  D. 


PROP.  XXII.    FIG.  14. 

IN  right  angled  spherical  triangles,  the  co-sine  of  either 
of  the  sides  is  to  the  radius,  as  the  co-sine  of  the  angle 
opposite  to  that  side  is  to  the  sine  of  the  other  angle. 

The  same  construction  remaining ;  in  the  triangle  CEF,  the 
sine  of  the  hypothenuse  CF  is  to  the  radius  as  the  sine  of  the 
side  EF  is  to  the  sine  of  the  angle  ECF  opposite  to  it ;  that  is, 
in  the  triangle  ABC,  the  co-sine  of  the  side  CA  is  to  the  radius, 
as  the  co-sine  of  the  angle  ABC  opposite  to  it,  is  to  the  sin.e  of 
the  other  angle.     Q.  E.  D. 


S  S 


i,06  SPHERICAL  TRIGONOMETRY. 


OF    THE   CIRCULAR  PARTS. 

Fig.  15.  IN  any  right  angled  spherical  triangle  ABC,  the  complement 
of  the  hypothenuse,  the  complements  of  the  angles,  and  the 
two  sides  are  called  The  circular  parts  of  the  triangle,  as  if  it 
■were  following;  each  other  in  a  circular  order,  from  whatever 
part  we  begin:  thus,  if  we  begin  at  the  complement  of  the 
hypothenuse,  and  proceed  towards  the  side  BA,  the  parts  fol- 
lowing in  order  will  be  the  complement  of  the  hypothenuse,  the 
complement  of  the  angle  B,  the  side  BA,  the  side  AC,  (for  the 
right  angle  at  A  is  not  reckoned  among  the  parts),  and,  lastly, 
the  complement  of  the  angle  C.  And  thus  at  whatever  part  we 
begin,  if  any  three  of  these  five  be  taken,  they  either  will  be 
all  contiguous  or  adjacent,  or  one  of  them  will  not  be  conti- 
guous to  either  of  the  other  two:  in  the  first  case,  the  part 
which  is  between  the  other  two  is  called  the  Middle  part^  and 
the  other  two  are  called  Adjacent  extremes.  In  llie  second  case, 
the  part  which  is  not  contiguous  to  either  of  the  other  two  i» 
called  the  Middle  part^  and  the  other  two  Opposite  extremes, 
I'or  example,  if  the  three  parts  be  the  complement  of  the  hy- 
pothenuse BC,  the  complement  of  the  angle  B,  and  the  sida 
BA  ;  since  these  three  are  contiguops  to  each  other,  the  com- 
plement of  the  angle  B  will  be  the  middle  part,  and  the  com- 
plement of  the  hypothenuse  BC  and  the  side  BA  will  be  adjacent 
extremes  :  but  if  the  complement  of  the  hypothenuse  BC,  and 
the  sides  BA,  AC  be  taken  ;  since  the  complement  of  the  hypo- 
thenuse is  not  adjacent  to  either  of  the  sides,  viz.  on  account 
of  the  complements  of  the  two  angles  B  and  C  intervening  be- 
tween it  and  the  sides,  the  complement  of  the  hypothenuse  BC 
will  be  the  middle  part,  and  the  sides,  BA,  AC  opposite  ex- 
tremes. The  most  acute  and  ingenious  Baron  Napier,  the  inven- 
tor of  Logarithms,  contrived  the  two  following  rules  concerning 
these  parts,  by  means  of  which  all  the  cases  of  right  angled 
spherical  triangles  are  resolved  with  the  greatest  ease. 

RULE  L 

The  rectangle  contained  by  the  radius  and  the  sine  of  the  mid- 
dle part,  is  equal  to  the  rectangle  contained  by  the  tangents  of 
the  adjacent  parts. 


SPHERICAL  TRIGONOMETRY.  507 


RULE  II. 

The  rectangle  contained  by  the  radius,  and  the  sine  of  the  mid- 
die  part  is  equal  to  the  rectangle  contained  by  the  co-sinea  of 
the  opposite  parts. 

These  rules  are  demonstrated  in  the  following  manner: 
First,  Let  either  of  the  sides,  as  BA,  be  the  middle  part,  and  Fij;.  ir- 
thereforc  the  complement  of  the  angle  B,  and  the  side  AC  will 
be  adjacent  extremes.     And  by  Cor.  2.  prop.  17.  of  this,  S,  BA 
is  to  the  Co-T,  B  as  T,  AC  is  to  the  radius,  and  therefore  RxS, 
BA=Co-T,  BxT,  AC. 

The  same  side  BA  being  the  middle  part,  the  complement  of 
the  hypothenuse,  and  the  complement  of  the  angle  C,  are  oppo- 
site ouremes ;  and  by  prop.  18.  S,  BC  is  to  the  radius,  as  S,  BA 
to  S,  C  ;  therefore  RxS,  BA=S,  BCxS,  C. 

Secondly,  Let  the  complement  of  one  of  the  angles,  as  B,  be 
the  middle  part,  and  the  complement  of  the  hypothenuse,  and 
the  side  BA  will  be  adjacent  extremes:  and  by  Cor.  prop.  20. 
Co-S,  B  is  to  Co-T,  BC,  as  T,  BA  is  to  the  I'adius,  and  therefore 
RxCo-S,  B=:Co-T,  BCxT,  BA. 

Again,  Let  the  complement  of  the  angle  B  be  the  middle  part, 
and  the  complement  of  the  angle  C,  and  the  side  AC  will  be  op- 
posite extremes :  and  by  prop.  22.  Co-S,  AC  is  to  the  radius,  as 
Co-S,  B  is  to  S,  C:  and  therefore  RxCo-S,  B=Co-S  ACxS,  C 
Thirdly,  Let  the  complement  of  the  hypothenuse  be  the  mid- 
dle part,  and  the  complements  of  the  angles  B,  C,  will  be  adja- 
cent extremes:  but  by  Cor.  .2.  prop.  19.  Co-S,  BC  is  to  Co-T, 
B  as  Co-T,  B  to  the  radius:  therefore  RxCo-S,  BC=Co-T 
CxCo-T,  C. 

Again,  Let  the  complement  of  the  hypothenuse  be  the  mid- 
dle part,  and  the  sides  AB,  AC  will  be  opposite  extremes :  but 
by  prop.  21.  Co-S,  AC  is  to  the  radius,  Co-S,  BC  to  Co-S,  BA  ; 
therefore  RxCo-S,  BC=Co.S,  BAxCo-S,  AC.     Q.  E.  D. 


5«8 


SPHERICAL  TRIGONOMETRY. 


SOLUTION   OF    THE    SIXTEEN    CASES    OF    RIGHT  ANGLE©  SPHE- 
RICAL   TRIANGLES. 

GENERAL  PROPOSITION. 

IN  a  right  angled  spherical  triangle,  of  the  three  sides 
and  three  angles,  any  two  being  given,  besides  the  right 
angle,  the  other  three  may  be  found. 

In  the  following  Table  the  solutions  are  derived  from  the  pre- 
ceding propositions.  It  is  obvious  that  the  same  solutions  may 
be  derived  from  Baron  Napier's  two  rules  above  demonstrated, 
which,  as  they  are  easily  remembered,  are  commonly  used  in 
practice. 


Case 

Given 

So't 

1 

AC,C 

B 

R  :  Co-S,  AC  :  :  b,  C  :  Co-S,  B  :  and  B  is 
ofthe  same  species  with  CA,  by  22.  and  13. 

2 

AC,  B 

c 

Co-S,  AC  :  R  :  :  Co-b,  B  :  S,  C  :  by  22. 

3 

B,  C 

AC 

S,  C  ;  Co-S,  B  :  :  R  :  Co-S,  AC  :  by  22. 
and  AC  is  of  the  same  species  with  B.  13. 

4 

BA,  AC 

BC 

R  :  Co-S,  BA  :  :  Co-S,  AC  :  Co-S,  BC. 
21.  and  if  both  BA,  AC  be  greater  or  less 
than  a  quadrant,  BC  will  be  less  than  a 
quadrant.  But  if  they  be  of  different  affec- 
tion, BC  will  be  greater  than  a  quadrant.  1 4. 

5 

BA,  BC 

AC 

Co-S,  B  A  :  R  :  :  Co-S,  BC  :  Co-S,  AC.  2 1 . 
and  if  BC  be  greater  or  less  than  a  quad- 
rant, BA,  AC  will  be  of  different  or  the 
same  affection:  by  15. 

6 

BA,  AC 

B 

S,  BA  :  R  :  :  T,  CA  :  T,  B.  17.  and  B  is 
of  the  same  affection  with  AC   13. 

#^ 


s 

s 

s 
s 
s 

\ 
\ 


SPHERICAL  TRIGONOMETRY. 


509 


^  Case 

S 
S 

S 

I 

s 
s 
s 
s 
s 


Given 


So't 


10 


BA,  B 


AC,  B 


AC 

"Ba 


BC,  C 


AC 


AC,  C 


BC 


R  :  b,  BA  :  :  T,  B  :  T,  Av^,  17.    And  AG 
is  of  the  same  affection  with  B.  13. 


T,  B  :  R  :  :  T,  CA  :  S,  BA.  17. 


R  :  Co-S,  C  :  :  T,  BC  :  T,  CA.  20.  If  BC 
be  less  or  greater  than  a  quadrant,  C  and 
B  will  be  of  the  same  or  different  affec- 
tion.  15.  13. 


S 

s 
s 

1 ,  BC  :  R  :  :  T,  CA  :  Co-is,  G.  20.  If  BC  \ 
be  less  or  greater  than  a  quadrant,  CA  and  S 
AB,  and  therefore  CA  and  C,  are  of  the  s 
same  or  different  affection.   15.  S 


Co-S,  C  :  R  :  :  T,  AG  :  1 ,  BC.  20.  And 
BC  is  less  or  greater  than  a  quadrant,  ac- 
cording as  C  and  AC  or  C  and  B  are  of  the 
same  or  different  affection.   14.  1. 


11 


12 


BC,CA 


C 


BC,  B 


AC 


K  :  b,  BC  :  :  S,  B  :  b,  AC.  18.  And  AG  ^ 
is  of  the  same  affection  with   B.  c 

S 

I 
\ 


\- 

S 

S' 

s 

s 


AC,  B 


bG 


b,  B  :  S,  AC  :  :  R  :  S,  BC  :  18. 


g    S,  BC  :  R  :  :  S,  AG  :  b,  B  :  18.  And  B  is 
of  the  same  affection  with  AC 


U 


BC,  AC 


15 


B,  C 


BC 


r,  C  :  R  :  :  Co-T,  B  :  Co-S,  BC.  19.  And 
according  as  the  angles  B  and  C  are  of  dif- 
ferent or  the  same  affection,  BC  will  be 
greater  or  less  than  a  quadrant.  14. 


16 


BC,  C 


B 


R  :  Go-b,  BC :  :  1  ,C  :  Go-T  ,B.  19.  If  BG 
be  less  or  greater  than  a  quadrant,  C  and  B 
will  be  of  the  same  ordifferentaffection.  15. 


519  SPHERICAL  TRIGONOMETRY. 

The  second,  eighth,  and  thirteenth  cases,  which  are  cominon- 
ly  called  ambiguous,  adnnit  of  two  solutions  :  for  in  these  it  is 
not  determined  whether  the  side  or  measure  of  the  angle  sought 
be  greater  or  less  than  a  quadrant. 


PROP.  XXIII.     FIG.  16. 

IN  spherical  triangles,  whetlier  right  angled  or  oblique 
angled,  the  sines  of  the  sides  are  proportional  to  the  sines 
of  the  angles  opposite  to  them. 

First,  let  ABC  be  a  right  angled  triangle,  having  a  right  an- 
gle at  A;  therefore  by  'prop.  18.  the  sine  of  the  hypothenuse  BC 
is  to  the  radius  (or  the  sine  of  the  right  angle  at  A)  as  the  sine 
of  the  side  AC  to  the  sine  of  the  angle  B.  And  in  like  manner, 
the  sine  of  BC  is  to  the  si;ie  of  the  angle  A,  as  the  sine  of  AB 
to  the  sine  of  the  angle  C  ;  wherefore  (11.  5.)  the  sine  of  the  side 
AC  is  to  the  sine  of  the  angle  B,  as  the  sine  of  AB  to  the  sine 
of  the  angle  C. 

Secondly,  let  BCD  be  an  oblique  angled  triangle,  the  sine  of 
either  of  the  sides  BC,  will  be  to  the  sine  of  either  of  the  other 
two  CD,  as  the  sine  of  the  angle  D  opposite  to  BC  is  to  the  sine 
of  the  angle  B  opposite  to  the  side  CD.  Through  the  point  C, 
let  there  be  drawn  an  arch  of  a  great  circle  CA  perpendicular 
upon  BD  ;  and  in  the  right  angled  triangle  ABC  (18.  of  this)  the 
sine  of  BC  is  to  the  i-adius,  as  the  sine  of  AC  to  the  sine  of  the 
angle  B;  and  in  the  triangle  ADC  (by  18.  of  this):  and,  by  in- 
version, the  radius  is  to  the  sine  of  DC  as  the  sine  of  the  angle 
D  to  the  sine  of  AC :  therefore  ex  xquo  perturbate,  the  sine  of 
BC  is  to  the  sine  of  DC,  as  the  sine  of  the  angle  D  to  the  sine 
of  the  angle  B.     Q.  E.  D. 


PROP.  XXIV.     FIG.  ir,  18. 

IN  oblique  angled  spherical  triangles  having  drawn  a 
perpendicular  arch  from  any  of  the  angles  upon  the  oppo- 
site side,  the  co-sines  of  the  angles  at  the  base  are  pro- 
portional to  the  sines  of  the  ^^ertical  angles. 

Let  BCD  be  a  triangle,  and  the  arch  CA  perpendicular  to  the 
base  BD  ;  the  co-sine  of  the  angle  B  will  be  to  the  co-sine  of 
the  angle  D,  as  the  sine  of  the  angle  BC  A  to  the  sine  of  the  an- 
trle  DCA. 


SPHERICAL  TRIGONOMETRY.  SIf 

For  by  22.  the  co-sine  of  the  angle  B  is  to  the  sine  of  the 
angle  BCA  as  (the  co-sine  of  the  side  AC  is  to  the  radius  ;  that 
is,  by  prop.  22.  as)  the  co-sine  of  the  angle  D  to  the  sine  of  the 
angle  DCA ;  and,  by  permutation,  the  co-sine  of  the  angle  R 
is  to  the  co-sine  of  the  angle  D,  as  the  sine  of  the  angle  BCA 
to  the  sine  of  the  angle  DCA.     Q.  E.  D. 

PROP.  XXV.    FIG.  17,  18. 

THE  same  things  remaining,  the  co-sines  of  the 
sides  BC,  CD,  are  proportional  to  tlie  co-sines  of  the 
bases  BA,  AD. 


For  by  21.  the  co-sine  of  BC  is  to  the  co-sine  of  BA,  as  (th* 
co-sine  of  AC  to  the  radius  ;  that  is,  by  21.  as)  the  co-sine  of 
CD  is  to  the  co-sine  of  AD  :  wherefore,  by  permutation*  the 
co-sines  of  the  sides  BC,  CD  are  proportional  to  the  co-sines  of 
the  bases  BA,  AD.     Q.  E.  D. 


PROP.  XXVI.     FIG.  ir,  18. 

THE  same  construction  remaining,  the  sines  of  the 
bases  BA,  AD  are  reciprocally  proportional  to  the  tan- 
gents of  the  angles  B  and  D  at  the  base. 

For  by  17.  the  sine  of  BA  is  to  the  radius,  as  the  tangent 
of  AC  to  the  tangent  of  the  angle  B ;  and  by  17.  and  inversion 
the  radius  is  to  the  sine  of  AD,  as  the  tangent  of  D  to  the  tan- 
gent of  AC  :  therefore,  ex  aquo  perturbate,  the  sine  of  BA  is 
to  the  sine  of  AD,  as  the  tangent  of  D  to  the  tangent  of  B. 

PROP.  XXVII.     FIG.  17,  IS. 

THE  CO- sines  of  the  vertical  angles  are  reciprocally 
proportional  to  the  tangents  of  the  sides. 

For  by  prop.  20.  the  co-sine  of  the  angle  BCA  is  to  the  ra- 
dius as  the  tangent  of  CA  is  to  the  tangent  of  BC  ;  and  by  the 
same  prop.  20^  and  by  inversion,  the  radius  is  to  the  co-sine  of 
the  angle  DCA,  as  the  tangent  of  DC  to  the  tangent  of  CA: 
therefore,  ex  aequo  perturbate,  the  co-sine  of  the  angle  BCA  is 


512  SPHERICAL  TRIGONOMETRY. 

to  the  ce^sine  of  the  angle  DCA,  as  the  tangent  of  DC  is  to  the 
tangent  of  BC.     Q.  E.  D. 


LEMMA.    FIG.  19,  20. 

IN  right  angled  plain  triangles,  the  hypotheniise  is  to 
the  radius,  as  the  excess  of  the  hypothenuse  above  either 
of  the  sides  to  the  versed  sine  of  the  acute  angle  adja- 
cent to  that  side,  or  as  the  sum  of  the  hypothenuse,  and 
either  of  the  sides  to  the  versed  sine  of  the  exterior 
angle  of  the  triangle. 

Let  the  triangle  ABC  have  a  right  angle  at  B  ;  AC  will  be 
to  the  radius  as  the  excess  of  AC  above  AB,  to  the  versed  sine 
of  the  angle  A  adjacent  to  AB ;  or  as  the  sum  of  AC,  AB  to  the 
versed  sine  of  the  exterior  angle  CAK. 

Vv'ith  any  radius  DE,  let  a  circle  be  described,  and  from  D 
the  centre  let  DF  be  drawn  to  the  circumference,  making  the 
angle  EDF  equal  to  the  angle  BAC,  and  from  the  point  F,  let 
FG  be  drawn  perpendicular  to  DE;  let  AH,  AK  be  made  equal 
to  AC,  and  DL  to  DE :  DG  therefore  is  the  co-sine  of  the  an- 
gle EDF  or  BAC,  and  GE  its  versed  sine :  and  because  of  the 
equiangular  triangles  ACB,  DFG,  AC  or  AH  is  to  DF  or  DE, 
as  AB  to  DG :  therefore  (19.  5.)  AC  is  to  the  radius  DE  as 
BH  toGE,  the  versed  sine  of  the  angle  EDF  or  BAC:  and  since 
AH  is  to  DE,  as  AB  to  DG  (12.  5.),  AH  or  AC  will  be  to  the 
radius  DE  as  KB  to  LG,  the  versed  sine  of  the  angle  LDF  or 
KAC.     Q.  E.  D. 


PROP.  XXVIII.     FIG.  21,  22. 

IN  any  spherical  triangle,  the  rectangle  contained  by 
the  sines  of  two  sides,  is  to  the  square  of  the  radius, 
as  the  excess  of  the  versed  sines  of  the  third  side  or 
base,  and  the  arch,  which  is  the  excess  of  the  sides  is 
to  the  versed  sine  of  the  angle  opposite  to  the  base. 

Let  ABC  be  a  spherical  triangle,  the  rectangle  contained  by 
the  sines  of  AB,  BC  will  be  to  the  square  of  the  radius,  as  the 
excess  of  the  versed  sines  of  the  base  AC,  and  of  the  arch,  which 
is  the  excess  of  AB,  BC  to  the  versed  sine  of  the  angle  ABC 
opposite  to  the  base. 


SPHERICAL  TRIGONOMETRY.  us 

Let  D  be  the  centre  of  the  sphere,  and  let  AD,  BD,  CD  be 
joined,  and  let  the  sines  AE,  CF,  CG  of  the  arches  AB,  BC, 
AC  be  drawn  ;  let  the  side  BC  be  greater  than  BA,  and  let  BH 
be  made  equal  to  BC  ;  AH  will  therefore  be  the  excess  of  the 
sides  BC,  BA  ;  let  HK  be  drawn  perpendicular  to  AD,  and 
since  AG  is  the  versed  sine  of  the  base  AC,  and  AK  the  versed 
sine  of  the  arch  AH,  KG  is  the  excess  of  the  versed  sines  of  the 
base  AC,  and  of  the  arch  AH,  which  is  the  excess  of  the  sides 
BC,  B  A :  let  GL  likewise  be  drawn  parallel  to  KH,  and  let  it 
meet  FH  in  L,  let  CL,  DH  be  joined,  and  let  AD,  FHmeet  each 
other  in  M. 

Since  therefore  in  the  triangles  CDF,  HDF,  DC,  DH  are  equal, 
DF  is  common,  and  the  angle  FDC  equal  to  the  angle  FDH, 
because  of  the  equal  arches  BC,  BH,  the  base  HF  will  be  equal 
to  the  base  FC,  and  the  angle  HFD  equal  to  the  right  angle  CFD  : 
the  straight  line  DF  therefore  (4.  11.)  is  at  right  angles  to  the 
plane  CFH :  wherefore  the  plane  CFH  is  at  right  angles  to  the 
plane  BDH,  which  passes  through  DF  (18.  11.).  In  like  man- 
ner, since  DG  is  at  right  angles  to  both  GC  and  GL,  DG  will  be 
perpendicular  to  the  plane  CGL  ;  therefore  the  plane  CGL  is  at 
right  angles  to  the  plane  BDH,  which  passes  through  DG :  and 
it  was  shown,  that  the  plane  CFH  or  CFL  was  perpendicular  to 
the  same  plane  BDH ;  therefore  the  common  section  of  the  planes 
CFL,  CGL,  viz.  the  straight  line  GL  is  perpendicular  to  the  plane 
BDA  (19.  11.),  and  therefore  CLF  is  a  right  angle:  in  the  trian- 
gle CFL  having  the  right  angle  CLF^  by  the  lemma  CF,  is  to 
the  radius  as  LH,  the  excesSf'vifT'df  CF  or  FH  above  EL,  is  to 
the  versed  sine  of  the  angle  CFL ;  but  the  angle  CFL  is  the  in- 
clination of  the  planes  BCD,  BAD,  since  FC,  FL  are  drawn  in 
them  at  right  angles  to  the  common  section  BF  :  the  spherical 
angle  ABC  is  therefore  the  same  with  the  angle  CFL  ;  and  there- 
fore CF  is  to  the  radius  as  LH  to  the  versed  sine  of  the  spherical 
angle  ABC  ;  and  since  the  triangle  AED  is  equiangular  (to  the 
triangle  MFD,  and  therefore)  to  the  triangle  MGL,  AE  will  be 
to  the  radius  of  the  sphere  AD,  (as  MG  to  ML  ;  that  is,  becaaSe 
of  the  parallels  as)  GK  to  LH  :  the  ratio  therefore  which  is  com- 
pounded of  the  ratios  of  AE  to  the  radius,  and  of  CF  to  the  same 
radius  ;  that  is,  (23.  6.)  the  ratio  of  the  rectangle  contained  by 
AE,  CF  to  the  square  of  the  radius,  is  the  same  with  the  ratio 
compounded  of  the  ratio  of  GK  to  LH,  and  the  ratio  of  LH  to 
the  versed  sine  of  the  angle  ABC  ;  that  is,  the  same  with  the 
ratio  of  GK  to  the  versed  sine  of  the  angle  ABC  ;  therefore,  the 
rectangle  contained  by  AE,  CF,  the  sines  of  the  sides  AB,  BC, 
is  to  the  square  of  the  radius  as  GK,  the  excess  of  the  versed 
sines  AG,.  AK,  of  the  base  AC,  and  the  arch  AH,  which  is  the 

3T 


SPHERICAL  TRIGONOMETRY. 

excess  of  the  sides  to  the  versed  sine  of  the  angle  ABC  opposite 
to  the  base  AC.     Q.  E.  D. 


PROP.  XXIX.     FIG.  23. 

THE  rectangle  contained  by  half  of  the  radius,  and 
the  excess  of  the  versed  sines  of  two  arches,  is  equal  to 
the  rectangle  contained  by  the  sines  of  half  the  sum,  and 
half  the  difference  of  the  same  arches. 

Let  AB,  AC  be  any  two  arches,  and  let  AD  be  made  equal  to 
AC  the  less  ;  the  arch  DB  therefore  is  the  sum,  and  the  arch  CB 
the  difference  of  AC,  AB  :  through  E  the  centre  of  the  circle^ 
let  there  be  drawn  a  diameter  DEF,  and  AE  joined,  and  CD 
likewise  perpendicular  to  it  in  G  ;  and  let  BH  be  perpendicular 
to  AE,  and  AH  will  be  the  versed  sine  of  the  arch  AB,  and  AG 
the  versed  sine  of  AC,  and  HG  the  excess  of  these  versed  sines: 
let  BD,  BC,  BF  be  joined,  and  FC  also  meeting  BH  in  K. 

Since  therefore  BH,  CG  are  parallel,  the  alternate  angles  BKC, 
KCG  will  be  equal  ;  but  KCG  is  in  a  semicircle,  and  therefore  a 
right  angle  ;  therefore  BKC  is  a  right  angle  ;  and  in  the  trian 
gles  DFB,  CBK,  the  angles  FDB,  BCK,  in  the  same  segment 
are  equal,  and  FBD,  BKC  are  right  angles ;  the  triangles  DFB, 
CBK  are  therefore  equiangular ;  wherefore  DF  is  to  DB,  as  BC 
to  CK,  or  HG  ;  and  therefore  the  rectangle  contained  by  the  di- 
ameter DF,  and  HG  is  equal  to  that  contained  by  DB,  BC  ; 
wherefore  the  I'ectangle  contained  by  a  fourth  part  of  the  diame- 
ter, and  HG,  is  equal  to  that  contained  by  the  halves  of  DB,  BC ; 
but  half  the  chord  DB  is  the  sine  of  half  the  arch  DAB,  that  is, 
half  the  sum  of  the  arches  AB,  AC  ;  and  half  the  chord  of  BC 
is  the  sine  of  half  the  arch  BC,  which  is  the  difference  of  ABj 
AC,     "Whence  the  proposition  is  manifest. 


PROP.  XXX.     FIG.  19,  24. 

THE  rectangle  contained  by  half  of  the  radius,  and 
the  versed  sine  of  any  arch,  is  equal  to  the  square  of  the 
sine  of  half  the  same  arch. 

Let  AB  be  an  arch  of  a  circle,  C  its  centre,  and  AC,  CB, 
BA  being  joined :  let  AB   be  bisectd*d  in  D,    and  let  CD  be 


SPHERICAL  TRIGONOMETRY.  516 

joined,  which  will  be  perpendicular  to  BA,  and  bisect  it  in  E 
(4.  !.).  BE  or  AE  therefore  is  the  sine  of  the  arch  DB  or  AD, 
the  half  of  AB :  let  BF  be  perpendicular  to  AC,  and  AF  will 
be  the  versed  sine  of  the  arch  BA :  but,  because  of  the  similar 
triangles  CAE,  BAF,  CA  is  to  AE,  as  AB,  that  is,  twice  AE,  to 
AF  ;  and  by  halving  the  antecedents,  half  of  the  radius  CA  is 
to  AE,  the  sine  of  the  arch  AD,  as  the  same  AE  to  AF  the  vers- 
ed sine  of  the  arch  AB.  Wherefore  by  16.  6.  the  proposition  is 
manifesJt. 


PROP.  XXXI.     FIG.  25. 

IN  a  spherical  triangle,  the  rectangle  contained  by  the 
sines  of  the  two  sides,  is  to  tlie  square  of  the  radius,  as 
the  rectangle  contained  by  the  sine  of  the  arch  which  is 
half  the  sum  of  the  base,  and  the  excess  of  the  sides,  and 
the  sine  of  the  arch,  which  is  half  the  difference  of  the 
same  to  the  square  of  die  sine  of  half  the  angle  opposite 
to  the  base. 

Let  ABC  be  a  spherical  triangle,  of  which  the  two  sides  are 
AB,  BC,  and  base  AC,  and  let  the  less  side  BA  be  produced,  so 
that  BD  shall  be  equal  to  BC  :  AD  therefore  is  the  excess  of  BC, 
BA;  and  it  is  to  be  shown,  that  the  rectangle  contained  by  the 
sines  of  BC,  BA  is  to  the  square  of  the  radius,  as  the  rectangle 
contained  by  the  sine  of  half  the  sum  of  AC,  AD,  and  the  sine 
of  half  the  difference  of  the  same  AC,  AD  to  the  square  of  the 
sine  of  half  the  angle  ABC,  opposite  to  the  base  AC. 

Since  by  prop.  28.  the  rectangle  contained  by  the  sines  of  the 
sides  BC,  BA  is  to  the  square  of  the  radius,  as  the  excess  of  the 
versed  sines  of  the  basse  AC  and  AD,  to  the  versed  sine  of  the 
angle  B;  that  is  (1.  6.),  as  the  rectangle  contained  by  half  the 
radius,  and  that  excess,  to  the  rectangle  contained  by  half  the 
radius,  and  the  versed  sine  of  B;  therefore  (29.  30.  of  this),  the 
recvangle  contained  by  the  sines  of  the  sides  BC,  BA  is  to  the 
square  of  the  radius,  as  the  rectangle  contained  by  the  sine  of 
the  arch,  which  is  half  the  sum  of  AC,  AD,  and  the  sine  of  the 
arch  which  is  half  the  difference  of  the  same  AC,  AD  is  to  the 
square  of  the  sine  of  half  the-  angle  ABC.     Q.  E.  D. 


516 


SPHERICAL  TRIGONOMETRY. 


SOLUTION    OF    THE  TWELVE  CASES  OF  OBLiqUE  ANGLED  SPHE- 
RICAL   TRIANGLES. 

GENERAL  PROPOSITION. 

IN  an  oblique  angled  spherical  triangle,  of  the  three 
sides  and  three  angles,  any  three  being  given,  the  other 
three  may  be  found. 


27. 


^..- 


Given 


B,  D,  and 
BC,  two  an- 
.^les  and  a 
side  opposite 
one  of  them. 


B,  C,  and 
3C,  two  an- 
gles and  tht 
side  betweer: 
ihem. 


BC,    CD 

and  B. 


BC,    DB, 
and  B. 


^' 


50  I 


C. 


D. 


aD 


CD 


Co-S,  BC  :  R  :  :  Co-T,  B  :  T,  BCA. 
19.  Likewise  by  24.  Co-S,  B  :  S,  BCA 
:  :  Co-S,  D  :  S,  DCA;  wherefore  BCD 
IS  the  sum  or  difference  of  the  angles 
DCA,  BCA  according  as  the  perpendi- 
cular CA  falls  within  or  without  the  tri- 
-ingle  BCD  ;  that  is  ('6.  of  this),  accord- 
ing as  the  angles  B,  D  are  of  the  same 
or  different  affection. 


Co-S,  BC  :  R  :  :  Co-T,  B  :  T,  BCA. 
19.  and  also  by  24.  S,  BCA  :  S,  DCA 
:  :  Co-S,  B  :  Co-S,  D  ;  and  according 
IS  the  angle  BCA  is  less  or  greater  than 
BCD,  the  perpendicular  CA  falls  within 
or  without  the  triangle  BCD  ;  and  there- 
fore (16.  of  this)  the  angles  B,  D  will 
be  of  the  same  or  different  affection. 


R  :  Co-S,  B  :  :  T,  BC  «  T,  BA.  20. 
ind  Co-S,  BC  :  Co-S,  BA  :  :  Co-S,  DC 
:  Co-S,  DA.  25.  and  BD  is  the  sum  or 
lifference  of  BA,  DA. 


R  :  Co-S,  B  :  :  T,  BC  ;  T,  BA.  20. 
uid  Co-S,  BA  :  Co-S,  BC  :  :  Co-S,  DA 
;  Co-S,  DC.  25.  and  according  as  DA, 
.\C  are  of  the  same  or  different  affec- 
ion  DC  will  be  less  or  greater  than  a 
juadrant.  14. 


"•^ 


SPHERICAL  TRIGONOMETRY. 


S\7 


S 


Given. 

B,  D  and 
BC. 

So't 
DB 

D. 

DC 

0. 

DC 

5 

R  :  Co-S,  B  I  :  T,  BC  :  T,  BA.  20. 
and  T,  D  :  T,  B  :  :  S,  BA  :  S,  DA.  26. 
and  BD  is  the  sum  or  difference  of  BA 
DA. 

6 

BC,   BD 

and  B. 

BC,   DC 
And  B. 

R  :  Co-S,  B  :  :  T,  BC  :  T,  BA.  20. 
.nd  S,  DA  :  S,  BA  :  :  T,  B  :  T,  D;  and 
according  as  BD  is  greater  or  less  than 
B  \,  the  angles  B,  D  are  of  the  same  or 
ilifferent  affection.  16. 

7 

Co-S,  BC  :  R  :  :  Co- T,  B  :  T,  BCA. 
19.  and  T,  DC  :  T,  BC  :  :  Co-S,  BCA  : 
Co-S,  DCA.  27.  the  sum  or  difference  of 
the  angles  BCA,  DCA  is  equal  to  the 
ingle  BCD. 

8 

B,  C  and 
BC. 

Co,S,  BC  :  R  :  :  Co-T,  B  :  T,  BCA. 
19.  also  by  27.  Co-S,  DCA  :  Co-S,  BCA 
:  :  T,  BC  :  T,  DC  27.  if  DCA  and  B 

■ic  of  the  same  affection;  that  is  (13.), 
.f  AD  and  CA  be  similar,  DC  will  be 
less  than  a  quadrant.  14.  and  if  AD,  CA 
be  not  of  the  same  affection,  DC  is  great- 
er than  a  quadrant.  14. 

9 

BC,  DC 

and  B. 

S,  CD  :  S,  B  :  :  S,  BC  :  S,  D. 

10 

B,  D  and 
BC. 

S,  D  :  S,  BC  :  :  S,  B  :  S,  DC. 

11 

BC,    BA 

AC. 
Fig.  25. 

B. 

S,  AB  X  S,  BC  :  R^^  :  :  S,  AC-f  AD 

2 

X  S,  AC— AD  :  Sg,  ABC.  See  Fig.  23. 

2                                       2 

AD  being  the  difference  of  the  sides  BC, 
BA. 

SPHERICAL  TRIGONOMETRY. 


S  12 

Given 

So't 

A,  B,  C. 

Th 

Fig.  7. 

sides. 

See  Fig.  7. 
In  the  triangles  DEF,  DE,  EF,  FD 
ire  respectively  the  supplements  of  the 
measures  of  the  given  angles  B,  A,  C 
in  the  triangle  BAG;  the  sides  of  the 
triangle  DEF  are  therefore  given,  and 
by  the  preceding  case  the  angles  D,  E, 
F  may  be  found,  and  the  sides  BC,  BA, 
AC  are  the  supplements  of  the  mea- 
sures of  these  angles. 


The  3d,  5th,  7th,  9th,  10th  cases,  which  are  commonly  called 
ambiguous,  admit  of  two  solutions,  either  of  which  will  answer 
the  conditions  required ;  for,  in  these  cases,  the  measure  of  the 
angle  or  side  sought,  may  be  either  greater  or  less  than  a  quad- 
rant, and  the  two  solutions  will  be  supplements  to  each  other 
<Cor.  to  def.  4.  6.  PI.  Tr.). 

If  from  any  of  the  angles  of  an  oblique  angled  spherical  trian- 
gle, a  perpendicular  arch  be  drawn  upon  the  opposite  side,  most 
of  the  cases  of  oblique  angled  triangles  may  be  resolved  by  means 
of  Napier's  rules. 


FINIS. 


M^.J. 


B 


Dl. 


D  G        B 


B 


Plate  11. 


Spherical.    Thigonome  try. 


Plate  11. 


Fig.  12 


A(- 


Fig. 


Flaxe     Trigonometry 


